ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.1

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ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.1

Question 1.
Classify the following measures as scalars and vectors:
(i) 5 seconds
(ii) 20 m/sec²
(iii) 1000 cm³
(iv) 2 metres north-west
(v) 2 km. (NCERT)
Solution:
(i) 5 seconds is a time so it is a scalar, since it has only magnitude.
(ii) 20 rn/sec² is acceleration, since it has both
magnitude and direction hence it is a vector quantity.
(iii) 1000 cm³ is volume so it has only magnitude and hence represents a scalar quantity.
(iv) 2 metres north-west is displacement so it has both magnitude and direction and hence represents a vector quantity.
(v) 40 watt is voltage so it has only magnitude and hence represents a scalar.
(vi) 2 km is distance so it has only magnitude and hence represents a scalar.

Question 1 (old).
(i) Represent graphically a displacement of 30 km, 30° east of north. (NCERT)
(ii) Represent graphically a displacement of 40 km, 30° west of south. (NCERT)
Solution:
(i) Here \(\overrightarrow{\mathrm{OP}}\) represents the displacement of 30 km, 30° east of north.

Scale 1 cm = 20 km

(ii) Here \(\overrightarrow{\mathrm{OP}}\) represents a displacement of 40 km, 30° west of south

Scale 1 cm = 20 km

Question 2.
The adjoining figure is a square. Identify the following vectors :
(i) Equal
(ii) Co-initial
(iii) Collinear but not equal (NCERT)

Solution:
(i) Two vectors \(\vec{a} \text { and } \vec{b}\) are equal if they have equal magnitude, same or parallel support and same direction.
∴ \(\vec{b} \text { and } \vec{d}\) are equal vectors, since all sides of square are equal.
(ii) vectors having same initial point are said to be co-initial vectors.
\(\vec{a} \text { and } \vec{d}\) are co-initial vectors.
(iii) Two or more vectors are said to be collinear iff they have same or parallel supports, irrespective of their magnitude and directions.
Clearly \(\vec{a} \text { and } \vec{c}\) are collinear but not equal vectors
while \(\vec{b} \text { and } \vec{d}\) are collinear and equal vectors.

Question 3.
State whether the following are true or false:
(i) Vectors \(\vec{a} \text { and }-\vec{a}\) a are collinear. (NCERT)
(ii) Two collinear vectors are always equal in magnitude. (NCERT)
(iii) Two vectors having same magnitude are collinear. (NCERT)
(iv) Two collinear vectors having same magnitude are equal. (NCERT)
(v) Zero vector is unique.
(vi) Modulus of vectors \(3 \vec{a} \text { and }-3 \vec{a}\) is same.
(vii) Resultant of \(\vec{a} \text { and }-\vec{a}\) is a proper vector.
(viii) Two equal and co-initial vectors must coincide.
(ix) (m + n) (\((\vec{a}+\vec{b})\)) = \(m \vec{a}+m \vec{b}+n \vec{a}+n \vec{b}\)
(x) (m n) \((\vec{a}+\vec{b})=m \vec{a}+n \vec{b}\)
(xi) \(\vec{a}=-\vec{b}\)
⇒ \(|\vec{a}|=|\vec{b}|\)
(xii) \(\vec{a}=\vec{b} \Rightarrow|\vec{a}|=|\vec{b}|\)
(xiii) \(|\vec{a}|=|\vec{b}| \Rightarrow \vec{a}=\vec{b}\)
(xiv) If \(\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{B C}}=\overrightarrow{0}\), then points O and C coincide.
Solution:
(i) True, since both vectors lies on a same line

(ii) False, since two or more vectors are said to be collinear iff they have same or || supports irrespective of their magnitudes and directions.

(iii) False, Since both vectors may have different supports.

(iv) False, Since both vectors may have different directions.

(v) False, \(\overrightarrow{\mathrm{AA}}, \overrightarrow{\mathrm{BB}}\) etc. are all coinitial vectors.

(vi) True, \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{BA}}\) are of same length i.e. \(|\overrightarrow{\mathrm{AB}}|=|\overrightarrow{\mathrm{BA}}|\)

(vii) False, since the resultant of \(\vec{a} \text { and }-\vec{a}\) is a zero vector.

(viii) True, Since coinitial vectors and equal vectors are of same magnitude.

(ix) True

(x) False, (m n) \((\vec{a}+\vec{b})=m n \vec{a}+m n \vec{b}\)

(xi) True, \(\vec{a}=-\vec{b}\)
⇒ \(|\vec{a}|=|-\vec{b}|=|\vec{b}|\)

(xii) True, \(\vec{a}=\vec{b}\)
⇒ \(|\vec{a}|=|\vec{b}|\)

(xiii) False,
⇒ \(\vec{a}=3 \hat{i}+4 \hat{j}\) ;
\(\vec{b}=4 i+3 \hat{j}\)
\(|\vec{a}|=|\vec{b}|=\sqrt{3^2+4^2}\) = 5 but \(\vec{a} \neq \vec{b}\)

(xiv) True, \(\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{0}\)
⇒ \(\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{0}\) [using ∆ law of vectors]
⇒ \(\overrightarrow{\mathrm{OC}}=\overrightarrow{0}\)
∴ \(\overrightarrow{\mathrm{O}} \text { and } \overrightarrow{\mathrm{C}}\) are coincide i.e. O and C are coincide.

Question 4.
If \(\vec{a} \text { and } \vec{b}\) are collinear, then state whether the following statements are true or false :
(i) b = λ . \(\vec{a}\) for some scalar.
(ii) \(\vec{a}= \pm \vec{b}\)
(iii) the respective components of \(\vec{a} \text { and } \vec{b}\) are proportional.
(iv) both the vectors \(\vec{a} \text { and } \vec{b}\) have same direction, but different magnitude.
Solution:
Since we know that, two or more vectors are collinear
iff they have same or parallel supports irrespective of their magnitudes and directions.

(i) True, since \(\vec{a} \text { and } \vec{b}\) are scalar multiple of each other.
∴ \(\vec{a}=\lambda \vec{b}\)
and \(\vec{b}=\mu \vec{a}\)
for some scalars λ and μ.

(ii) false, since \(\vec{a} \text { and } \vec{b}\) are collinear
∴ \(\vec{a}=\lambda \vec{b}\) for some scalar λ
∴ \(\vec{a} \neq \pm \vec{b}\) for λ ≠ ± 1

(iii) True, Let \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\)
and \(\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\)
Since \(\vec{a}=\lambda \vec{b}\) for some scalar λ.
∴ \(\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right)=\lambda\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)\)
⇒ a₁ = λ b₁ ;
a₂ = λ b₂ ;
a₃ = λ b₃
Thus their components are proportional

(iv) False, Since \(\vec{a}=\lambda \vec{b}\) for some scalar λ.
Since \(\vec{a}=-2 \vec{b}\) for λ = – 2,
Here \(\vec{a} \text { and } \vec{b}\) are of opposite direction but \(\vec{a} \text { and } \vec{b}\) are collinear.

Question 5.
(i) If k \(\vec{a}=\overrightarrow{0}\), then what can you say about k and \(\vec{a}\) ?
(ii) If \(\vec{a}\) is a non-zero vector, then find a scalar k such that |k \(\vec{a}\)| = 1.
Solution:
(i) k \(\vec{a}\) = \(\overrightarrow{0}\)
⇒ k = 0 or \(\vec{a}=\overrightarrow{0}\)
or k and \(\vec{a}\) both are \(\overrightarrow{0}\).

(ii)

Question 5 (old).
In a triangle ABC (shown in the adjoining figure), state whether the following are true or false :
(i) \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
(ii) \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{B C}}-\overrightarrow{\mathbf{A C}}=\overrightarrow{0}\)
(iii) \(\overrightarrow{\mathbf{A B}}-\overrightarrow{\mathbf{C B}}+\overrightarrow{\mathbf{C A}}=\overrightarrow{0}\)
(iv) \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{B C}}-\overrightarrow{\mathbf{C A}}=\overrightarrow{0}\)

Solution:
(i) By ∆ law of addition of vectors
\(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}\)
⇒ \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}\) ……………..(1)
⇒ \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
∴ given statement is true.

(ii) From eqn. (1) ; given result is true.

(iii) True, Since \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
⇒ \(\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
[∵ \(\overrightarrow{\mathrm{BC}}=-\overrightarrow{\mathrm{CB}}\)]

(iv) False, since \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)

Question 6.
If \(|\vec{a}|\) = 5, then calculate
(i) |2 \(\vec{a}\)|
(ii) \(\left|-\frac{1}{2} \vec{a}\right|\)
(iii) \(-\frac{1}{2}|\vec{a}|\)
Solution:
Given \(|\vec{a}|\) = 5
(i) \(|2 \vec{a}|\) = |2| \(|\vec{a}|\)
= 2 × 5 = 10

(ii) \(\left|-\frac{1}{2} \vec{a}\right|=\left|-\frac{1}{2}\right||\vec{a}|\)
= \(\frac{1}{2}\) × 5
= \(\frac{5}{2}\)

(iii) \(-\frac{1}{2}|\vec{a}|\)
= – \(\frac{1}{2}\) × 5
= – \(\frac{5}{2}\)

Question 7.
(i) For what value of p, is \((\hat{i}+\hat{j}+\hat{k})\) p a unit vector.
(ii) If \(\vec{a}=-3 \hat{i}-2 \hat{j}+6 \hat{k}\), then find the value of λ so that λ \(\vec{a}\) may be a unit vector.
Solution:
(i) Let \(\vec{a}\) = p \((\hat{i}+\hat{j}+\hat{k})\)
= \(p \hat{i}+p \hat{j}+p \hat{k}\)
Since \(\vec{a}\) be a unit vector.
∴ \(|\vec{a}|\) = 1
⇒ \(\sqrt{p^2+p^2+p^2}\) = 1
⇒ \(\sqrt{3 p^2}\) = 1
⇒ √3 p = 1
⇒ √3 |p| = 1
⇒ |p| = \(\frac{1}{\sqrt{3}}\)
⇒ p = ± \(\frac{1}{\sqrt{3}}\)

(ii) Given \(\vec{a}=-3 \hat{i}-2 \hat{j}+6 \hat{k}\)
Since λ \(\vec{a}\) is a unit vector

Question 8.
(i) Find values of x and y so that the vectors \(2 \hat{i}+3 \hat{j} \text { and } x \hat{i}+y \hat{j}\) are equal. (NCERT)
(ii) Find the values of x, y and z so that the vectors \(\vec{a}=x \hat{i}+2 \hat{j}+z \hat{k}\) and \(\vec{b}=2 \hat{i}+y \hat{j}+\hat{k}\) are equal. (NCERT)
(iii) If \(\vec{a}=x \hat{i}+2 \hat{j}-z \hat{k}\) and \(\vec{b}=3 \hat{i}-y \hat{j}+\hat{k}\) are two equal vectors, then write the value of x + y + z.
(iv) If \(\vec{a}=x \hat{i}+2 \hat{j}-3 \hat{k}\) and \(\vec{b}=3 \hat{i}-y \hat{j}+z \hat{k}\), then find the values of x, y and z so that \(2 \vec{a}=3 \vec{b}\).
Solution:
(i) Let \(\vec{a}=2 \hat{i}+3 \hat{j}\)
and \(\vec{b}=x \hat{i}+y \hat{j}\)
Since \(\vec{a} \text { and } \vec{b}\) are equal.
∴ \(\vec{a}=\vec{b}\)
⇒ \(2 \hat{i}+3 \hat{j}=x \hat{i}+y \hat{j}\)
⇒ x = 2 and y = 3.
[on comparing the coefficients of i and j on both sides]

(ii) Since \(\vec{a}=\vec{b}\)
⇒ \(x \hat{i}+2 \hat{j}+z \hat{k}=2 \hat{i}+y \hat{j}+\hat{k}\)
on comparing the coefficients of \(\hat{i}, \hat{j} \text { and } \hat{k}\) on both sides; we have
x = 2 ; y = 2 ; z = 1

(iii) Given \(\vec{a}=x \hat{i}+2 \hat{j}-z \hat{k}\)
and \(\vec{b}=3 \hat{i}-y \hat{j}+\hat{k}\)
Now \(\vec{a}=\vec{b}\)
⇒ \(x \hat{i}+2 \hat{j}-z \hat{k}=3 \hat{i}-y \hat{j}+\hat{k}\)
∴ x = 3 ; y = – 2 and z = – 1.
Thus x + y + z = 3 – 2 – 1 = 0

(iv) Given \(\vec{a}=x \hat{i}+2 \hat{j}-3 \hat{k}\)
and \(\vec{b}=3 \hat{i}-y \hat{j}+z \hat{k}\)
Since \(2 \vec{a}=3 \vec{b}\)
⇒ \(2(x \hat{i}+2 \hat{j}-3 \hat{k})=3(3 \hat{i}-y \hat{j}+z \hat{k})\)
⇒ \(2 x \hat{i}+4 \hat{j}-6 \hat{k}=9 \hat{i}-3 y \hat{j}+3 z \hat{k}\)
[since two vectors are equal iff their corresponding components are equal]
2x = 9 ;
4 = – 3y
and – 6 = 3z
⇒ x = \(\frac{9}{2}\) ;
y = – \(\frac{4}{3}\)
and z = – 2

Question 9.
(i) Let \(\vec{a}=\hat{i}+2 \hat{j}\) and \(\vec{b}=2 \hat{i}+\hat{j}\). Is \(|\vec{a}|=|\vec{b}|\) ? Are the vectors \(\vec{a} \text { and } \vec{b}\) equal ? (NCERT)
(ii) Write two different vectors having same magnitude. (NCERT)
(iii) Write two different vectors having same direction. (NCERT)
(iv) If \(|\vec{a}|=|\vec{b}|\), is it true that \(\vec{a}= \pm \vec{b}\) ? Justify your answer.
Solution:
(i) Given \(\vec{a}=\hat{i}+2 \hat{j}\)
and \(\vec{b}=2 \hat{i}+\hat{j}\)
Here \(|\vec{a}|=\sqrt{1^2+2^2}=\sqrt{5}\)
and \(|\vec{b}|=\sqrt{2^2+1^2}=\sqrt{5}\)
∴ \(|\vec{a}|=|\vec{b}|\)
Clearly \(\vec{a} \text { and } \vec{b}\) are not equal.

(ii) Let \(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}\) ;
\(|\vec{a}|=\sqrt{2^2+(-1)^2+2^2}\) = 3
& \(\vec{b}=-2 \hat{i}+\hat{j}-2 \hat{k}\) ;
\(|\vec{b}|=\sqrt{(-2)^2+1^2+(-2)^2}=\sqrt{9}\) = 3

(iii) \(\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k}\)
& \(\vec{b}=4 \hat{i}+2 \hat{j}+4 \hat{k}\)
Clearly \(\vec{b}=2 \vec{a}\)
i. e. \(\vec{b} \text { and } \vec{a}\) have same direction.
Since both \(\vec{a} \text { and } \vec{b}\) are parallel vectors.

(iv) Let \(\vec{a}=3 \hat{i}+4 \hat{j}\)
⇒ \(|\vec{a}|=\sqrt{3^2+4^2}\) = 5
and \(\vec{b}=4 \hat{i}+3 \hat{j}\)
⇒ \(|\vec{b}|=\sqrt{4^2+3^2}\) = 5
∴ \(|\vec{a}|=|\vec{b}|\)
but \(\vec{a} \neq \vec{b}\) and \(\vec{a} \neq-\vec{b}\)

Question 10.
If a unit vector \(\vec{a}\) makes angles \(\frac{\pi}{3}\) with \(\hat{i}, \frac{\pi}{4} \text { with } \hat{j}\) and an acute angle θ with \(\hat{k}\), then find the value of θ.
Solution:

Question 11.
Find the magnitude of the following vectors :
(i) \(\hat{i}+\hat{j}+\hat{k}\)
(ii) \(2 \hat{i}-3 \hat{j}-7 \hat{k}\)
(iii) \(\frac{1}{\sqrt{3}} \hat{i}-\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}\) (NCERT)
Solution:
(i) Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\)
∴ \(|\vec{a}|=\sqrt{1^2+1^2+1^2}\)
= \(\sqrt{1+1+1}=\sqrt{3}\)

(ii) Let \(\vec{a}=2 \hat{i}-3 \hat{j}-7 \hat{k}\)
∴ \(|\vec{a}|=|2 \hat{i}-3 \hat{j}-7 \hat{k}|\)
= \(\sqrt{2^2+(-3)^2+(-7)^2}\)
= \(\sqrt{4+9+49}=\sqrt{62}\)

(iii) Let \(\vec{a}=\frac{1}{\sqrt{3}} \hat{i}-\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}\)
∴ \(|\vec{a}|=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+\left(-\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2}\)
= \(\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}\)
= \(\sqrt{\frac{3}{3}}=\sqrt{1}\) = 1

Question 12.
(i) Find the vector with initial point P (2, 3, 0) and terminal point Q (- 1, – 2, – 4). Also find its magnitude. (NCERT)
(ii) Write the scalar components of the vector \(\overrightarrow{\mathrm{AB}}\) with initial point A (2, 1) and terminal point B (- 5, 7).
(iii) Find the terminal point of the vector PQ whose initial point is P (- 2, 5, 0) and components along coordinate axes are 2, – 3 and 7 respectively.
Solution:
(i)

(ii) P.V of A = \(2 \hat{i}+\hat{j}\)
P.V of B = \(-5 \hat{i}+7 \hat{j}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \(-5 \hat{i}+7 \hat{j}-(2 \hat{i}+\hat{j})\)
⇒ \(\overrightarrow{\mathrm{AB}}=-7 \hat{i}+6 \hat{j}\)
∴ Scalar components of vector \(\overrightarrow{\mathrm{AB}}\) are – 7 and 6.

(iii) Given, P.V of initial point P = \(-2 \hat{i}+5 \hat{j}+0 \hat{k}\)
Let Q be the terminal point of \(\overrightarrow{\mathrm{PQ}}\) with P.V \(\vec{b}\).
Since \(\overrightarrow{\mathrm{PQ}}\) is having components along coordinate axes are 2, – 3 and 7.
Thus, \(\overrightarrow{\mathrm{PQ}}\)
⇒ P.V of Q – P.V of P = \(2 \hat{i}-3 \hat{j}+7 \hat{k}\)
⇒ \(\vec{b}-(-2 \hat{i}+5 \hat{j}+0 \hat{k})=2 \hat{i}-3 \hat{j}+7 \hat{k}\)
⇒ \(\vec{b}=(2 \hat{i}-3 \hat{j}+7 \hat{k})+(-2 \hat{i}+5 \hat{j}+0 \hat{k})\)
⇒ \(\vec{b}=0 \hat{i}+2 \hat{j}+7 \hat{k}\)
Hence the terminal point of \(\overrightarrow{\mathrm{PQ}}\) be Q(0, 2, 7).

Question 13.
(i) Find the sum of the vectors \(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}\), \(\vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}\).
(ii) If \(\vec{a}, \vec{b}, \vec{c}\) have components (1, – 1), (2, – 2) and (2, 1) respectively, find \(2 \vec{a}-\vec{b}+3 \vec{c}\).
Solution:
(i) Given \(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}\),
\(\vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}\)
and \(\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}\)
∴ \(\vec{a}+\vec{b}+\vec{c}\) = \((1-2+1) \hat{i}+(-2+4-6) \hat{j}+(1+5-7) \hat{k}\)
= \(0 \hat{i}-4 \hat{j}-\hat{k}\)

(ii) Since \(\vec{a}, \vec{b} \text { and } \vec{c}\) have components (1, – 1), (2, – 2) and (2, 1).

Question 14.
Find a unit vector in the direction of the vector \(\vec{a}=2 \hat{i}-3 \hat{j}\).
(ii) Find a unit vector in the direction of the vector \(\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k}\).
(iii) Find a unit vector in the direction of the vector \(\vec{a}=3 \hat{i}-2 \hat{j}+6 \hat{k}\).
Solution:
(î) required unit vector in the direction of \(\vec{a}=2 \hat{i}-3 \hat{j}\)
= \(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{2 \hat{i}-3 \hat{j}}{|2 \hat{i}-3 \hat{j}|}=\frac{2 \hat{i}-3 \hat{j}}{\sqrt{2^2+(-3)^2}}\)
= \(\frac{2}{\sqrt{13}} \hat{i}-\frac{3}{\sqrt{13}} \hat{j}\)

(ii) Given \(\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k}\)
∴ \(|\vec{b}|=\sqrt{2^2+1^2+2^2}\) = 3
Thus required unit vector in the direction of \(\vec{b}\)
\(\hat{b}=\frac{\vec{b}}{|\vec{b}|}\)
= \(\frac{2 \hat{i}+\hat{j}+2 \hat{k}}{3}\)
= \(\frac{2}{3} \hat{i}+\frac{1}{3} \hat{j}+\frac{2}{3} \hat{k}\)

(iii) Given \(\vec{a}=3 \hat{i}-2 \hat{j}+6 \hat{k}\)
∴ \(|\vec{a}|=\sqrt{3^2+(-2)^2+6^2}=\sqrt{49}\) = 7
Thus required unit vector in the direction of \(\vec{a}\)
= \(\hat{a}=\frac{\vec{a}}{|\vec{a}|}\)
= \(\frac{3 \hat{i}-2 \hat{j}+6 \hat{k}}{7}\)
= \(\frac{3}{7} \hat{i}-\frac{2}{7} \hat{j}+\frac{6}{7} \hat{k}\)

Question 15.
(i) Find a unit vector in the direction of \(\) where P and Q have coordinates (5, 0, 8) and (3, 3, 2) respectively. (NCERT Exemplar)
(ii) Find the unit vector in the direction of AB where A and B are the points (1, 2, 3) and (4, 5, 6) respectively. (NCERT)
Solution:
(i) Here \(\overrightarrow{\mathrm{PQ}}\) = P.V of Q – P.V of P
= \((3 \hat{i}+3 \hat{j}+2 \hat{k})-(5 \hat{i}+0 \hat{j}+8 \hat{k})\)
= \(-2 \hat{i}+3 \hat{j}-6 \hat{k}\)
∴ required unit vector in the direction of \(\overrightarrow{\mathrm{PQ}}=\hat{\mathrm{PQ}}\)
= \(\frac{\overrightarrow{P Q}}{|\overrightarrow{P Q}|}\)
= \(\frac{-2 \hat{i}+3 \hat{j}-6 \hat{k}}{\sqrt{(-2)^2+3^2+(-6)^2}}\)
= \(\frac{-2 \hat{i}+3 \hat{j}-6 \hat{k}}{\sqrt{4+9+36}}\)
= \(-\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}-\frac{6}{7} \hat{k}\)

(ii) \(\overrightarrow{\mathrm{OA}}\) = P.V. of A
= \(\hat{i}+2 \hat{j}+3 \hat{k}\)
and \(\overrightarrow{\mathrm{OB}}\) = P.V. of B
= \((4 \hat{i}+5 \hat{j}+6 \hat{k})\)
∴ \(\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}\) = P.V. of B – P.V. of A
= \((4 \hat{i}+5 \hat{j}+6 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})\)
= \(3 \hat{i}+3 \hat{j}+3 \hat{k}\)
∴ \(\hat{\mathrm{AB}}=\frac{\overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{AB}}|}=\frac{3(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3^2+3^2+3^2}}\)
= \(\frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}\)

Question 16.
(i) Write a unit vector in the direction of sum of vectors \(\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}\) and \(\vec{b}=2 \hat{i}+\hat{j}-7 \hat{k}\).
(ii) If \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{b}=2 \hat{i}+\hat{j}-2 \hat{k}\), find the unit vector in the direction of \(2 \vec{a}-\vec{b}\).
Solution:
(i) Given \(\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}\)
and \(\vec{b}=2 \hat{i}+\hat{j}-7 \hat{k}\)

(ii) Given \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k}\)
and \(\vec{b}=2 \hat{i}+\hat{j}-2 \hat{k}\)

Question 17.
(i) Find a unit vector parallel to te sum of vectors \(\vec{a}=2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\). (NCERT)
(ii) If \(\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}\) and \(\vec{b}=2 \hat{i}+4 \hat{j}+9 \hat{k}\), then find a unit vector parallel to \(\vec{a}+\vec{b}\).
Solution:
(i) Given \(\vec{a}=2 \hat{i}+4 \hat{j}-5 \hat{k}\)
and \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\)

(ii) Given \(\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}\)
and \(\vec{b}=2 \hat{i}+4 \hat{j}+9 \hat{k}\)

Question 18.
(i) Find a vector in the direction of the vector \(\vec{a}=\hat{i}-2 \hat{j}\), whose magnitude is 7 units.
(ii) Write a vector of magnitude 9 units in the direction of the vector \(-2 \hat{i}+\hat{j}+2 \hat{k}\).
(iii) Find vectors of magnitude 5 units which are parallel to the sum of vectors \(\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}-2 \hat{j}+\hat{k}\).
Solution:
(i) Given \(\vec{a}=\hat{i}-2 \hat{j}\)
∴ required vector in the direction of \(\vec{a}\) having magnitude 7 units.
= \(7 \hat{a}=\frac{7 \vec{a}}{|\vec{a}|}\)
= \(\frac{7(\hat{i}-2 \hat{j})}{\sqrt{1^2+(-2)^2}}\)
= \(\frac{7}{\sqrt{5}}(\hat{i}-2 \hat{j})\)

(ii) We required a vector of magnitude 9 which is in the direction of \(\vec{a}=-2 \hat{i}+\hat{j}+2 \hat{k}\)
i.e. required vector = \(\frac{9 \vec{a}}{|\vec{a}|}\)
= \(\frac{9(-2 \hat{i}+\hat{j}+2 \hat{k})}{\sqrt{(-2)^2+1^2+2^2}}\)
= \(3(-2 \hat{i}+\hat{j}+2 \hat{k})\)

(iii) Given \(\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}\) ;
and \(\vec{b}=\hat{i}-2 \hat{j}+\hat{k}\)
∴ \(\vec{a}+\vec{b}=3 \hat{i}+\hat{j}+0 \hat{k}\)
\(|\vec{a}+\vec{b}|=\sqrt{3^2+1^2+0^2}=\sqrt{10}\)
Thus, required unit vector of magnitude 5
= \(\frac{ \pm 5(\vec{a}+\vec{b})}{|(\vec{a}+\vec{b})|}\)
= \(\frac{ \pm 5(3 \hat{i}+\hat{j})}{\sqrt{10}}\)

Question 18 (old).
(i) If \(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}\) and \(\vec{b}=-\hat{i}+\hat{j}-\hat{k}\), then find the unit vector in the direction of the vector \(\vec{a}+\vec{b}\). (NCERT)
(ii) If \(\vec{a}=\hat{i}+\hat{j}\), \(\vec{b}=\hat{j}+\hat{k}\) and \(\vec{c}=\hat{k}+\hat{i}\) then find a unit vector in the direction of \(\vec{a}+\vec{b}+\vec{c}\).
Solution:
(i) Given \(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}\)
and \(\vec{b}=-\hat{i}+\hat{j}-\hat{k}\)
∴ \(\vec{a}+\vec{b}=(2 \hat{i}-\hat{j}+2 \hat{k})+(-\hat{i}+\hat{j}-\hat{k})\)
= \(\hat{i}+0 \hat{j}+\hat{k}\)
⇒ \(\vec{c}=\vec{a}+\vec{b}\)
= \(\hat{i}+0 \hat{j}+\hat{k}\)
Thus unit vector in the direction \(\vec{a}+\vec{b}\)
∴ \(\hat{c}=\frac{\vec{c}}{|\vec{c}|}\)
= \(\frac{\hat{i}+0 \hat{j}+\hat{k}}{\sqrt{1^2+1^2}}\)
= \(\frac{1}{\sqrt{2}}(\hat{i}+\hat{k})\)

(iv) Given \(\vec{a}=\hat{i}+\hat{j}\) ;
\(\vec{b}=\hat{j}+\hat{k}\) ;
and \(\vec{c}=\hat{k}+\hat{i}\)

Question 19.
(i) If the magnitude of the position vector of the point (3, – 2, α) is 7 units, then find all possible values of α.
(ii) What vector should be added to the vector \(\vec{a}=3 \hat{i}+4 \hat{j}-2 \hat{k}\) so that the resultant is a unit vector \(\hat{i}\)?
Solution:
(i) Let \(\vec{a}\) be the position vector of point (3, – 2, α)
∴ \(\vec{a}=3 \hat{i}-2 \hat{j}+\alpha \hat{k}\) also given \(|\vec{a}|\) = 7
⇒ \(\sqrt{3^2+(-2)^2+\alpha^2}\) = 7
⇒ \(\sqrt{9+4+\alpha^2}\) = 7
⇒ \(\sqrt{13+\alpha^2}\) = 7
Oh squaring; we have
13 + α² = 49
⇒ α² = 49 – 13 = 36
⇒ α = ± 6

(ii) Given \(\vec{a}=3 \hat{i}+4 \hat{j}-2 \hat{k}\)
Let the required vector be \(\vec{b}\) so that
\(\vec{a}+\vec{b}=\hat{i}\)
⇒ \(\vec{b}=\hat{i}-\vec{a}\)
= \(\hat{i}-(3 \hat{i}+4 \hat{j}-2 \hat{k})\)
⇒ \(\vec{b}=-2 \hat{i}-4 \hat{j}+2 \hat{k}\)

Question 20.
(i) Show that the vectors \(2 \hat{i}-3 \hat{j}+4 \hat{k}\) and \(-4 \hat{i}+6 \hat{j}-8 \hat{k}\) are collinear.
(ii) Write the value of p for which \(\vec{a}=3 \hat{i}+2 \hat{j}+9 \hat{k}\) and \(\vec{b}=\hat{i}+p \hat{j}+3 \hat{k}\) are parallel vectors.
(iii) If \(\vec{a}=p \hat{i}+3 \hat{j}\) and \(\vec{b}=4 \hat{i}+p \hat{j}\), then find the values of p so that \(\vec{a} \text { and } \vec{b}\) may be parallel.
Solution:
(i) Let O be the origin and A, B be the points whose position vectors are

Thus \(\overrightarrow{\mathrm{OA}} \text { and } \overrightarrow{\mathrm{OB}}\) vectors are parallel vectors and the origin O is common to them.
Thus \(\overrightarrow{\mathrm{OA}} \& \overrightarrow{\mathrm{OB}}\) vectors are collinear,
Therefore A, B are collinear vectors.

(ii) Given \(\vec{a}=3 \hat{i}+2 \hat{j}+9 \hat{k}\)
\(\vec{b}=\hat{i}+p \hat{j}+3 \hat{k}\) are parallel vectors
∴ \(\vec{a}=\lambda \vec{b}\) for some non-zero scalar λ.
⇒ \(3 \hat{i}+2 \hat{j}+9 \hat{k}=\lambda(\hat{i}+p \hat{j}+3 \hat{k})\)
∴ 3 = λ ;
2 = λ p
and 9 = 3λ
Thus, 2 = 3p
⇒ p = \(\frac{2}{3}\)

(iii) Given \(\vec{a}=p \hat{i}+3 \hat{j}\)
and \(\vec{b}=4 \hat{i}+p \hat{j}\)

Question 20 (old).
(iii) Find the vector in the direction of the vector \(\vec{r}=5 \hat{i}-\hat{j}+2 \hat{k}\) which has magnitude 8 units.
(iv) Find vectors of magnitude 5 units which are parallel to the vector \(3 \hat{i}+\hat{j}\).
Solution:
(iii) Let \(\vec{a}=5 \hat{i}-\hat{j}+2 \hat{k}\),
Let < l, m, n > be the direction cosines of \(\vec{a}\)
The direction ratios of \(\vec{a}\) are proportional to < 5, – 1, 2 >
∴ direction cosines of \(\vec{a}\) are \(\left\langle\frac{5}{\sqrt{5^2+(-1)^2+2^2}}, \frac{-1}{\sqrt{5^2+(-1)^2+2^2}}, \frac{2}{\sqrt{5^2+(-1)^2+2^2}}\right\rangle\)
i.e. \(\left\langle\frac{5}{\sqrt{30}}, \frac{-1}{\sqrt{30}}, \frac{2}{\sqrt{30}}\right\rangle\)
required vector = 8 \((\ell \hat{i}+m \hat{j}+n \hat{k})\)
= 8 \(\left(\frac{5}{\sqrt{30}} \hat{i}-\frac{1}{\sqrt{30}} \hat{j}+\frac{2}{\sqrt{30}} \hat{k}\right)\)
= \(\frac{8}{\sqrt{30}}(5 \hat{i}-\hat{j}+2 \hat{k})\)

(iv) Given \(\vec{a}=3 \hat{i}+\hat{j}\)
So a unit vector parallel to \(\vec{a}= \pm \hat{a}\)
= \(\pm \frac{\vec{a}}{|\vec{a}|}\)
= \(\pm \frac{(3 \hat{i}+\hat{j})}{\sqrt{3^2+1}}\)
= \(\pm \frac{(3 \hat{i}+\hat{j})}{\sqrt{10}}\)
required vector = \(5 \hat{a}= \pm \frac{5}{\sqrt{10}}(3 \hat{i}+\hat{j})\)

Question 21.
(i) Write the direction ratios of the vector \(\hat{i}+\hat{j}-2 \hat{k}\), and hence calculate its direction cosines.
(ii) Find the direction cosines of the vector joining the points A (1, 2, – 3) and B (- 1, 2, 1), directed from A to B.
Solution:
(i) Let \(\vec{a}=\hat{i}+\hat{j}-2 \hat{k}\)
∴ D ratios of \(\vec{a}\) be < 1, 1, – 2 >
Thus, D’ cosines of \(\vec{a}\) are
< \(\frac{1}{\sqrt{1^2+1^2+(-2)^2}}, \frac{1}{\sqrt{1^2+1^2+(-2)^2}}, \frac{-2}{\sqrt{1^2+1^2+(-2)^2}}\) >
i.e., < \(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}\) >

(ii) direction ratios of vector \(\overrightarrow{\mathrm{AB}}\) are ;
< – 1 – 1 , – 2 – 2, 1 + 3 >
i.e., < – 2, – 4, 4 >
i.e., < – 1 , – 2, 2 >
Thus direction cosines of vector \(\overrightarrow{\mathrm{AB}}\) are ;
< \(\frac{-1}{\sqrt{(-1)^2+(-2)^2+2^2}}, \frac{-2}{\sqrt{(-1)^2+(-2)^2+2^2}}, \frac{2}{\sqrt{(-1)^2+(-2)^2+2^2}}\) >
i.e., < \(-\frac{1}{3},-\frac{2}{3}, \frac{2}{3}\) >

Question 22.
(i) Find the mid – point of the vector joining the points P and Q having position vectors \(2 \hat{i}+3 \hat{j}-4 \hat{k}\) and \(4 \hat{i}+\hat{j}-2 \hat{k}\) respectively.
(ii) The position vector of the mid-point M of the line segment AB is \(3 \hat{i}+4 \hat{k}\). If the position vector of the point A is \(4 \hat{i}-2 \hat{j}+3 \hat{k}\), find the position vector of B.
Solution:
(i) Given P.V. of P = \(2 \hat{i}+3 \hat{j}-4 \hat{k}\)
and P.V. of Q = \(4 \hat{i}+\hat{j}-2 \hat{k}\)

∴ P.V. of mid-point of line joining P and Q = \(\frac{(2 \hat{i}+3 \hat{j}-4 \hat{k})+(4 \hat{i}+\hat{j}-2 \hat{k})}{2}\)
= \(\frac{6 \hat{i}+4 \hat{j}-6 \hat{k}}{2}\)
= \(3 \hat{i}+2 \hat{j}-3 \hat{k}\)

(ii) Let P.V. of B = \(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}\)
given P.V. of A = \(4 \hat{i}-2 \hat{j}+3 \hat{k}\)

Question 22 (old).
(ii) Write the position vector of the point dividing the line segment joining the points with position vector \(\vec{a} \text { and } \vec{b}\) externally in the ratio 1 : 4, where \(\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}\) and \(\vec{b}=-\hat{i}+\hat{j}+\hat{k}\)
Solution:
Given \(\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}\)
and \(\vec{b}=-\hat{i}+\hat{j}+\hat{k}\)
∴ \(\vec{a}=\lambda \vec{b}\)
⇒ \(3 \hat{i}+2 \hat{j}+9 \hat{k}=\lambda(\hat{i}+p \hat{j}+3 \hat{k})\)
∴ 3 = λ ;
2 = λ p
and 9 = 3λ
Thus, 2 = 3p
⇒ p = \(\frac{2}{3}\)

Question 23 (old).
(ii) Find the direction cosines of the vector \(\hat{i}+2 \hat{j}+3 \hat{k}\). (NCERT)
(iii) If P (1, 5, 4) and Q (4, 1, – 2), then find the direction ratios of \(\overrightarrow{\mathrm{PQ}}\).
Solution:
(ii) Let \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\)
∴ direction ratio of are proportional to < 1, 2, 3 >
Thus, direction cosines of \(\vec{a}\) are ;
\(\left\langle\frac{1}{\sqrt{1^2+2^2+3^2}}, \frac{2}{\sqrt{1^2+2^2+3^2}}, \frac{3}{\sqrt{1^2+2^2+3^2}}\right\rangle\)
i.e., \(\left\langle\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right\rangle\)

(iii) We know that,
If P (x₁, y₁, z₁) and Q (x₂, y₂, z₂)
Then D’ ratios of \(\overrightarrow{\mathrm{PQ}}\) are
< x₂ – x₁, y₂ – y₁, z₂ – z₁ >
∴ D’ ratios of \(\overrightarrow{\mathrm{PQ}}\) are < 4 – 1, 1 – 5, – 2 – 4 >
i.e., < 3, – 4, – 6 >
Thus, D’ cosines of \(\overrightarrow{\mathrm{PQ}}\) are
< \(\frac{3}{\sqrt{3^2+(-4)^2+(-6)^2}}, \frac{-4}{\sqrt{3^2+(-4)^2+(-6)^2}}, \frac{-6}{\sqrt{3^2+(-4)^2+(-6)^2}}\) >
i.e., < \(\frac{3}{\sqrt{9+16+36}}, \frac{-4}{\sqrt{9+16+36}}, \frac{-6}{\sqrt{9+16+36}}\) >
i.e., < \(\frac{3}{\sqrt{61}}, \frac{-4}{61}, \frac{-6}{61}\) >

Question 23.
If the position vectors of theverticesA, B and C of a ∆ABC are \(\hat{i}+2 \hat{j}-3 \hat{k}\), \(-2 \hat{i}-5 \hat{j}+4 \hat{k}\) and \(7 \hat{i}-4 \hat{k}\) respectively, then write the position vector of the centrold of ∆ABC.
Solution:
Let \(\vec{a}, \vec{b}, \vec{c}\) be the position vectors of the vertices A, B and C of ∆ABC.
Then, \(\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k\)k ;
\(\vec{b}=-2 \hat{i}-5 \hat{j}+4 \hat{k\)
and \(\vec{c}=7 \hat{i}-4 \hat{k}\)
Thus the P.V of centroid of ∆ABC are \(\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)\)
= \(\frac{(\hat{i}+2 \hat{j}-3 \hat{k})+(-2 \hat{i}-5 \hat{j}+4 \hat{k})+(7 \hat{i}-4 \hat{k})}{3}\)
= \(\frac{6 \hat{i}-3 \hat{j}-3 \hat{k}}{3}\)
= \(2 \hat{i}-\hat{j}-\hat{k}\)

Question 24.
Find the position vector of a point R which divides the line segment joining the points P and Q with position vectors \(\) and \(\) respectively in the ratio 2 : 1
(i) internally
(ii) externally. (NCERT)
Solution:
(i) P.V. of point R
= \(\frac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2 \hat{j}-\hat{k})}{2+1}\)
= \(\frac{(-2+1) \hat{i}+(2+2) \hat{j}+(2-1) \hat{k}}{2+1}\)
= \(-\frac{\hat{i}}{3}+\frac{4}{3} \hat{j}+\frac{1}{3} \hat{k}\)

(ii) P.V. of poiont R = \(\frac{2(-\hat{i}+\hat{j}+\hat{k})-1(\hat{i}+2 \hat{j}-\hat{k})}{2-1}\)
= \((-2-1) \hat{i}+(2-2) \hat{j}+(2+1) \hat{k}\)
= \(-3 \hat{i}+3 \hat{k}\)

Question 25.
(i) Find the position vector of the point which divides the join of the points with position vectors \(\vec{a}+3 \vec{b}\) and \(\vec{a}-\vec{b}\) internally in the ratio 1 : 3.
(ii) X and Y are two points with position vectors \(3 \vec{a}+\vec{b}\) and \(\vec{a}-3 \vec{b}\) respectively. Write the position vector of a point Z which divides the line segment XY in the ratio 2 : 1 externally.
Solution:
(i) Then by using section formula, P.V. of point which divides the line segment AB internally in the ratio 1 : 3

(ii) Thus by section formula.
The P.V of point which divides the line segment AB externally in the ratio 2: 1 i.e. 2 : – 1 internally

Question 26.
The vectors a and b are non-zero non-collinear. For what value of x, the vectors (x – 2) \(\) and (2x + 1) \(\) are collinear?
Solution:
Let \(\vec{c}=(x-2) \vec{a}+\vec{b}\)
and \(\vec{d}=(2 x+1) \vec{a}-\vec{b}\)
Since \(\vec{c}=\lambda \vec{d}\) are collinear.
∴ \(\vec{c} \text { and } \vec{d}\)
for some non-zero scalar λ.
⇒ (x – 2) \(\vec{a}+\vec{b}\) = λ [(2x + 1) \(\vec{a}-\vec{b}\)]
Since \(\vec{a} \text { and } \vec{b}\) are non-zero non-collinear vectors.
∴ x – 2 = λ (2x + 1).
and 1 = – λ
⇒ λ = – 1
∴ from (1) ;
x – 2 = λ (2x + 1)
⇒ x – 2 = – 2x – 1
⇒ 3x = 1
⇒ x = \(\frac{1}{3}\)

Question 27.
The position vectors of four points A, B, C, D are \(\vec{a}, \vec{b}, 2 \vec{a}+3 \vec{b}, 2 \vec{a}-3 \vec{b}\) respectively. Express the vectors \(\overrightarrow{\mathbf{A C}}, \overrightarrow{\mathbf{B D}}, \overrightarrow{\mathbf{C D}}, \overrightarrow{\mathbf{A B}}\) in terms of \(\vec{a} \text { and } \vec{b}\).
Solution:

Question 27 (old) .
(i) A and B are two points with position vectors \(2 \vec{a}-3 \vec{b}\) and \(6 \vec{b}-\vec{a}\) respectively. Write the position vector of a point P which divides the line segment AB internally in the ratio 1 : 2.
(ii) P and Q are points with position vectors \(3 \vec{a}-2 \vec{b}\) and \(\vec{a}+\vec{b}\) respectively. Write the position vector of a point R which divides the line segment PQ in the ratio 2: 1 externally.
Solution:
(i) required P.V of a point P which divides AB internally in the ratio 1 : 2 is given by

(ii) The P.V R of point which divides the line segment externally in the ratio 2 : 1.
i.e. internally 2 : – 1
\(\overrightarrow{\mathrm{R}}=\frac{2(\vec{a}+\vec{b})-(3 \vec{a}-2 \vec{b})}{2-1}\)
= \(-\vec{a}+4 \vec{b}\)

Question 28.
(i) If the points A, B, C and D have coordinates (0, 1), (1, 0), (1, 2) and (2, 1) respectively, then prove that \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{CD}}\).
(ii) If the coordinates of the points A and B in a plane are (1, 1) and (2, 2) respectively, then find the coordinates of point C such that \(\overrightarrow{\mathbf{A B}}=\overrightarrow{\mathrm{BC}}\).
Solution:
(i) Let \(\vec{a}, \vec{b}, \vec{c} \text { and } \vec{d}\) are the position vectors of points A, B, C and D respectively.

(ii) Given P.V of A = \(\hat{i}+\hat{j}\)
P.V of B = \(2 \hat{i}+2 \hat{j}\)
Let the coordinates of C be (a, b)
∴ P.V. of C = \(a \hat{i}+b \hat{j}\)
Since \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{BC}}\)
⇒ P.V of B – P.V of A = P.V of C – P.V of B
⇒ \((2 \hat{i}+2 \hat{j})-(\hat{i}+\hat{j})=(a \hat{i}+b \hat{j})-(2 \hat{i}+2 \hat{j})\)
⇒ \(\hat{i}+\hat{j}=(a-2) \hat{i}+(b-2) \hat{j}\)

Since if two vectors are equal then their corresponding components are also equal.
∴ 1 = a – 2
⇒ a = 3
and 1 = b – 2
⇒ b = 3.
Hence the coordinates of point C be (3, 3).

Question 29.
If the components of \(\vec{a}\) are (1, 2) and \(\vec{b}=\hat{i}-\hat{j}+\vec{a}\), then what are the componcnts of \(2 \vec{a}-3 \vec{b}\) ?
Solution:
Since the components of \(\vec{a}\) are (1, 2)
∴ \(\vec{a}=\hat{i}+2 \hat{j}\)
and \(\vec{b}=\hat{i}-\hat{j}+\vec{a}\)
= \(\hat{i}-\hat{j}+\hat{i}+2 \hat{j}\)
= \(2 \hat{i}+\hat{j}\)
∴ \(2 \vec{a}-3 \vec{b}=2(\hat{i}+2 \hat{j})-3(2 \hat{i}+\hat{j})\)
= \(2 \hat{i}+4 \hat{j}-6 \hat{i}-3 \hat{j}\)
= \(-4 \hat{i}+\hat{j}\)
∴ Components of \(2 \vec{a}-3 \vec{b}\) are (- 4, 1).

Question 30.
Let \(\overrightarrow{\mathrm{PQ}}\) be a vector of magnitude 8 units making an angle of 30° with x-axis lying in fourth quadrant. Find its components along x-axis and y-axis.
Solution:
Given |\(\overrightarrow{\mathrm{PQ}}\)| = 8 units
and \(\overrightarrow{\mathrm{PQ}}\) making an angle of 30° with x-axis and lying in fourth quadrant.
Let < l, m > be the direction cosine of \(\overrightarrow{\mathrm{PQ}}\)
∴ l = cos 330°
= cos (360° – 30°)
= cos 30°
= \(\frac{\sqrt{3}}{2}\)
and l² + m² = 1
m² = 1 – l²
m = – \(\sqrt{\left(1-\frac{3}{4}\right)}\)
= \(-\sqrt{\frac{1}{4}}=-\frac{1}{2}\)

or m = cos (240°)
= cos (180° + 60°)
= – cos 60°
= – \(\frac{1}{2}\)
∴ |\(\overrightarrow{\mathrm{PQ}}=|\overrightarrow{\mathrm{PQ}}|(l \hat{i}+m \hat{j})\)|
= 8 \(\left(\frac{\sqrt{3}}{2} \hat{i}-\frac{1}{2} \hat{j}\right)\)
⇒ \(\overrightarrow{\mathrm{PQ}}=4 \sqrt{3} \hat{i}-4 \hat{j}\)
Thus, components of \(\overrightarrow{\mathrm{PQ}}\) along x – axis and y – axis are ; 4√3 and – 4.

Question 31.
(i) Show that the points (0, – 1), (1, 0) and (3, 2) are collinear.
(ii) The position vectors of the points A, B, C are \(2 \hat{i}+\hat{j}-\hat{k}, 3 \hat{i}-2 \hat{j}+\hat{k}\) and \(\hat{i}+4 \hat{j}-3 \hat{k}\) respectively, show that A, B, C are collinear.
Solution:
(i) The position vectors of points A (0, – 1), B (1, 0) and C (3, 2) are \(-\hat{j}, \hat{i}, 3 \hat{i}+2 \hat{j}\).
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \(\hat{i}-(-\hat{j})\)
= \(\hat{i}+\hat{j}\)

∴ \(\overrightarrow{\mathrm{AC}}\) = P.V. of C – P.V. of A
= \(3 \hat{i}+2 \hat{j}-(-\hat{j})\)
= \(3 \hat{i}+3 \hat{j}\)

Since \(\overrightarrow{\mathrm{AB}}=\frac{1}{3} \overrightarrow{\mathrm{AC}}\)

Thus the vectors \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}}\) are collinear.

∴ \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}}\) are having same or parallel supports.

But \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}}\) are coinitial vectors

∴ they are having same support. Thus, the points A, B and C are collinear.

(ii) Let A, B and Care the points whose position vectors are \(2 \hat{i}+\hat{j}-\hat{k}\), \(3 \hat{i}-2 \hat{j}+\hat{k}\) and \(\hat{i}+4 \hat{j}-3 \hat{k}\)

Thus \(\overrightarrow{\mathrm{BC}} \text { and } \overrightarrow{\mathrm{AB}}\) are parallel vectors and the point B is common to both vectors.
Thus \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{BC}}\) are collinear vectors and hence the points A, B, and c are collinear points.

Question 32.
If \(\vec{a}, \vec{b}\) are position vectors of points A (1, – 1) and B (- 3, m), then find the value of m so that the origin O, and points A and B are
(i) collinear
(ii) not collinear.
Solution:
(i) Given \(\vec{a}\) = P.V.of A
= \(\hat{i}-\hat{j}\)
and \(\vec{b}\) = P.V.of B
= \(-3 \hat{i}+m \hat{j}\)
Since the points O, A and B are collinear.
Thus \(\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}}\) are collinear.
∴ \(\overrightarrow{\mathrm{OA}}=\lambda \overrightarrow{\mathrm{OB}}\) for sorne non-zero scalar λ.
⇒ \(\vec{a}-\overrightarrow{0}=\lambda(\vec{b}-\overrightarrow{0})\)
⇒ \(\hat{i}-\hat{j}=\lambda(-3 \hat{i}+m \hat{j})\)
∴ 1 = – 3 λ
and – 1 = λ m
⇒ λ = – \(\frac{1}{3}\)
and – 1 = – \(\frac{1}{3}\) m
m = 3.

(ii) Now points O, A, B are non collinear if m ≠ 3.

Question 33.
(i) if the points (α, – 1), (2, 1) and (4, 5) are collinear, then find a by vector method.
(ii) Using vectors, find he value of λ such that (λ, – 10, 3), (1, – 1, 3) amd (3, 5, 3) are collinear. (NCERT Exemplar)
Solution:
(i) Let the given points are A (α, – 1), B (2, 1) and C (4, 5)
∴ P.V of A = \(\alpha \hat{i}-\hat{j}\) ;
P.V of B = \(2 \hat{i}+\hat{j}\)
and P.V of C = \(4 \hat{i}+5 \hat{j}\)

∴ \(\overrightarrow{\mathrm{AB}}\) = P.V of B – P.V of A
= \((2 \hat{i}+\hat{j})-(\alpha \hat{i}-\hat{j})\)
= (2 – α) \(\hat{i}+2 \hat{j}\)

\(\overrightarrow{\mathrm{AC}}\) = P.V of C – P.V of A
= \((4 \hat{i}+5 \hat{j})-(\hat{i}-\hat{j})\)
= (4 – α) \(\hat{i}+6 \hat{j}\)

Since A, B and C are collinear points.
Thus \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}}\) are collinear vectors.
Thus, \(\overrightarrow{\mathrm{AB}}=\lambda \overrightarrow{\mathrm{AC}}\) for some non-zero scalar λ.

⇒ \((2-\alpha) \hat{i}+2 \hat{j}\) = λ \([(4-\alpha) \hat{i}+6 \hat{j}]\)
∴ 2 – α = λ (4 – α)
and 2 = 6λ
λ = \(\frac{1}{3}\)
⇒ 2 – α = \(\frac{1}{3}\) (4 – α)
⇒ 6 – 3α = 4 – α
⇒ 3α – α = 6 – 4
⇒ 2α = 2
⇒ α = 1

(ii) Let A (λ, – 10, 3), B (1, – 1, 3) and C (3, 5, 3) are given points.
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V.of B – P.V.of A
= \((\hat{i}-\hat{j}+3 \hat{k})-(\lambda \hat{i}-10 \hat{j}+3 \hat{k})\)
= \((1-\lambda) \hat{i}+9 \hat{j}+0 \hat{k}\)

& \(\overrightarrow{\mathrm{BC}}\) = P.V.of C – P.V.of B
= \((3 \hat{i}+5 \hat{j}+3 \hat{k})-(\hat{i}-\hat{j}+3 \hat{k})\)
= \(2 \hat{i}+6 \hat{j}+0 \hat{k}\)
Since A, B, C are collinear points.
∴ \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{BC}}\) are collinear vectors.
∴ \(\overrightarrow{\mathrm{AB}}=m \overrightarrow{\mathrm{BC}}\) for some scalar m.
⇒ \((1-\lambda) \hat{i}+9 \hat{j}=m[2 \hat{i}+6 \hat{j}]\)
Comparing the coefficients of \(\hat{i} \& \hat{j}\) on both sides, we have,
1 – λ = 2m ………………….(1)
and 9 = 6m
⇒ m = 3/2
∴ From (1) ; we have
1 – λ = 2 × \(\frac{3}{2}\)
⇒ 1 – λ = 3
⇒ λ = – 2

Question 33. (old).
(ii) Show that the points A (2, 6, 3), B (1, 2, 7) and C (3, 10, – 1) are collinear.
Solution:
Let O be the origin of reference

∴ \(\overrightarrow{\mathrm{AC}} \text { and } \overrightarrow{\mathrm{AB}}\) are parallel and also coinitial vectors.
Hence A, B, C are collinear.

Question 34.
If \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\), \(\vec{b}=4 \hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{c}=\hat{i}-2 \hat{j}+\hat{k}\) then find a vector of magnitude 6 units which is parallel to the vector \(2 \vec{a}-\vec{b}+3 \vec{c}\).
Solution:
Given \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\),
\(\vec{b}=4 \hat{i}-2 \hat{j}+3 \hat{k}\)
and \(\vec{c}=\hat{i}-2 \hat{j}+\hat{k}\)

Question 35.
(i) Show that the points A, B and C with position vectors \(3 \hat{i}-4 \hat{j}-4 \hat{k}\), \(2 \hat{i}-\hat{j}+\hat{k}\) and \(\hat{i}-3 \hat{j}-5 \hat{k}\) respectively form the vertices of a right-angled triangle. (NCERT)
(ii) If the vertices of a triangle are A (2, – 1, 1), B (1, – 3, – 5) and C (3, – 4, – 4), then prove by vector method that it is a right- angled triangle.
(iii) If the position vectors of the vertices of a triangle ABC are \(\hat{i}+2 \hat{j}+3 \hat{k}\), \(2 \hat{i}+3 \hat{j}+\hat{k}\) and \(3 \hat{i}+\hat{j}+2 \hat{k}\), then prove that ∆ABC is an equilateral triangle.
Solution:
Let A, B, C are the points whose position vectors are \(\vec{a}, \vec{b}, \vec{c}\)
i.e., \(\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}\) ;
\(\vec{b}=2 \hat{i}-\hat{j}+\hat{k}\)
and \(\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}\)

Clearly BC² = AB² + AC²
[∵ 41 = 35 + 6]
Hence pythagoras theorem is verified.
Thus the points A, B, C form the vertices of right angled ∆.

(ii) Let A, B, C are the points whose position vectors are latex]\vec{a}, \vec{b}, \vec{c}[/latex].

So points A, B and C form a triangle.
Clearly
AB² = CA² + CB²
[∵ 41 = 35 + 6]
Hence pythagoras theorem is verified. Thus the points A, B C form the vertices of right angled ∆.

(iii) Let O be the origin of reference.

Question 36.
The position vectors \(\vec{a}, \vec{b}, \vec{c}\) of three given points satisfy the relation \(4 \vec{a}-9 \vec{b}+5 \vec{c}=\overrightarrow{0}\). Prove that the three points are collinear.
Solution:
Let \(\vec{a}, \vec{b}, \vec{c}\) are position vectors of given points A, B and C respectively.
Since \(4 \vec{a}-9 \vec{b}+5 \vec{c}=\overrightarrow{0}\)
⇒ \(9 \vec{b}=4 \vec{a}+5 \vec{c}\)
⇒ \(\vec{b}=\frac{4 \vec{a}+5 \vec{c}}{4+5}\)

Thus point B divides CA internally in the ratio 4 : 5.
∴ point B lies on CA.
Thus A, B, C lies on same line and hence the given points A, B and C are collinear.

Question 38 (old).
If \(\vec{a}, \vec{b}\) are position vectors of A, B with reference to origin O, find the position vector of a point C in AB produced such that 2AC = 3AB.
Solution:
Given \(\vec{a}, \vec{b}\) are the position vectors of A
and B with reference to origin O.
∴ \(\overrightarrow{\mathrm{OA}}=\vec{a}\) ;
\(\overrightarrow{\mathrm{OB}}=\vec{b}\)
Given, 2AC = 3AB
⇒ \(2 \overrightarrow{\mathrm{AC}}=3 \overrightarrow{\mathrm{AB}}\)
⇒ \(2(\vec{c}-\vec{a})=3(\vec{b}-\vec{a})\)
where \(\vec{c}\) be the position vectors of point C.
⇒ \(2 \vec{c}=3 \vec{b}-3 \vec{a}+2 \vec{a}=3 \vec{b}-\vec{a}\)
⇒ \(\vec{c}=\frac{3 \vec{b}-\vec{a}}{2}\)