The availability of step-by-step OP Malhotra ISC Class 12 Solutions Chapter 21 Vectors Ex 21(c) can make challenging problems more manageable.
S Chand Class 12 ICSE Maths Solutions Chapter 21 Vectors Ex 21(c)
Question 1.
If the position vector of a point (-4, -3) be \(\vec{a}\), find |\(\vec{a}\)|.
Answer:
Given \(\vec{a}\) = -4 \(\hat{i}\) – 3 \(\hat{j}\)
⇒ |\(\vec{a}\)| = \(\sqrt{(-4)^2+(-3)^2}\)
= \(\sqrt{16+9}\)
= \(\sqrt{25}\) = 5
Question 2.
If P(-1, 3) and Q(2, -7) express the vector \(\vec{r}\) = \(\overrightarrow{\mathrm{PQ}}\) in terms of unit vectors \(\hat{i}\) and \(\hat{j}\). Also determine the magnitude of \(\hat{r}\).
Answer:
Given P.V of P= \(-\hat{i}\) + 3 \(\hat{j}\)
P.V of Q = 2 \(\hat{i}\) – 7 \(\hat{j}\)
∴ \(\vec{r}\) = \(\overrightarrow{\mathrm{PQ}}\) = P.V of Q – P.V of P
= (2 \(\hat{i}\) – 7 \(\hat{j}\)) – (\(-\hat{i}\) + 3 \(\hat{j}\)) = 3 \(\hat{i}\) – 10 \(\hat{j}\)
∴ |\(\vec{r}\)| = \(\sqrt{3^2+(-10)^2}\)
= \(\sqrt{9+100}\)
= \(\sqrt{109}\)
Question 3.
I \(\vec{a}\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) and \(\vec{b}\) = \(-\hat{i}\) + \(\hat{j}\), find the magnitude of 2 \(\vec{a}\) – 3 \(\vec{b}\).
Answer:
Given \(\vec{a}\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) and \(\vec{b}\) = \(-\hat{i}\) + \(\hat{j}\)
2 \(\vec{a}\) – 3 \(\vec{b}\) = 2(3 \(\hat{i}\) – 2 \(\hat{j}\)) – 3(\(-\hat{i}\) + \(\hat{j}\))
= 9 \(\hat{i}\) – 7 \(\hat{j}\)
∴ |2 \(\vec{a}\) – 3 \(\vec{b}\)| = |9 \(\hat{i}\) – 7 \(\hat{j}\)|
= \(\sqrt{9^2+(-7)^2}\) = \(\sqrt{81+49}\) = \(\sqrt{130}\)
Question 4.
If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) have components (1, 1) (2, 3) and (-1, 4) respectively, find the components of 2 \(\vec{a}\) – 3 \(\vec{b}\) + 5 \(\vec{c}\).
Answer:
Given \(\vec{a}\) = \(\hat{i}\) + \(\hat{j}\) ;
\(\vec{b}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) and \(\vec{c}\) = \(-\hat{i}\) + 4 \(\hat{j}\)
∴ 2 \(\vec{a}\) – 3 \(\vec{b}\) + 5 \(\vec{c}\)
= 2(\(\hat{i}\) + \(\hat{j}\)) – 3(2 \(\hat{i}\) + 3 \(\hat{j}\)) + 5(\(-\hat{i}\) + 4 \(\hat{j}\))
= 2 \(\hat{i}\) + 2 \(\hat{j}\) – 6 \(\hat{i}\) – 9 \(\hat{j}\) – 5 \(\hat{i}\) + 20 \(\hat{j}\)
= -9 \(\hat{i}\) + 13 \(\hat{j}\)
Thus the components of 2 \(\vec{a}\) – 3 \(\vec{b}\) + 5 \(\vec{c}\) are (-9, 13).
Question 5.
Find a unit vector parallel to 3 \(\hat{i}\) + 4 \(\hat{j}\).
Answer:
Let \(\vec{a}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\)
∴|\(\vec{a}\)| = \(\sqrt{3^2+4^2}\) = \(\sqrt{25}\) = 5
∴ unit vector parallel to \(\vec{a}\)
= ± \(\hat{a}\) = ± \(\frac{\vec{a}}{|\vec{a}|}\)
= ± \(\frac{(3 \hat{i}+4 \hat{j})}{5}\)
Question 6.
Find a unit vector in the direction of \(\hat{i}\) + \(\hat{j}\).
Answer:
Let \(\vec{a}\) = \(\hat{i}\) + \(\hat{j}\)
∴ |\(\vec{a}\)| = \(\sqrt{1^2+1^2}\) = \(\sqrt{2}\)
So unit vector in the dirction of \(\vec{a}\)
= \(\hat{a}\) = \(\frac{\vec{a}}{|\vec{a}|}\)
= \(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\)
Question 7.
Find a vector of magnitude 5 units which is parallel to the vector 2 \(\hat{i}\) – \(\hat{j}\).
Answer:
Let \(\vec{a}\) = 2 \(\hat{i}\) – \(\hat{j}\)
∴|\(\vec{a}\)| = \(\sqrt{2^2+(-1)^2}\) = \(\sqrt{5}\)
∴ unit vector parallel to the direction of \(\vec{a}\)
= ± \(\hat{a}\)
= ± \(\frac{\vec{a}}{|\vec{a}|}\)
Thus, vector of magnitude 5 units which is || to \(\vec{a}\)
= ± 5 \(\hat{a}\)
= ± 5 \(\frac{\vec{a}}{|\vec{a}|}\)
= ± 5 \(\frac{(2 \hat{i}-\hat{j})}{\sqrt{5}}\)
= ± \(\sqrt{5}\)(2 \(\hat{i}\) – \(\hat{j}\))
Question 8.
If the components of \(\vec{a}\) are (1, 2) and \(\vec{b}\) = \(\hat{i}\) – \(\hat{j}\) + \(\vec{a}\), what are components of 2 \(\vec{a}\) – 3 \(\vec{b}\) ?
Answer:
Given \(\vec{a}\) = \(\hat{i}\) + 2 \(\hat{j}\)
and
\(\vec{b}\) = \(\hat{i}\) – \(\hat{j}\) + \(\vec{a}\)
= \(\hat{i}\) – \(\hat{j}\) + \(\hat{i}\) + 2 \(\hat{j}\)
= 2 \(\hat{i}\) + \(\hat{j}\)
Thus, 2 \(\vec{a}\) – 3 \(\vec{b}\) = 2(\(\hat{i}\) + 2 \(\hat{j}\)) – 3(2 \(\hat{i}\) + \(\hat{j}\))
= -4 \(\hat{i}\) + \(\hat{j}\)
Therefore the components of 2 \(\vec{a}\) – 3 \(\vec{b}\) are (-4, 1).
Question 9.
Find the terminal point of the vector \(\overrightarrow{P Q}\) whose initial points is P(1, 2) and whose components along x-axis and y-axis are -2 and 3 respectively.
Answer:
Let the terminal point of vector \(\overrightarrow{P Q}\) be \(\vec{Q}\) and has coordinates (α, β) i.e.
P.V of Q = α \(\hat{i}\) + β \(\hat{j}\)
and
P.V of P = \(\hat{i}\) + 2 \(\hat{j}\)
\(\overrightarrow{\mathrm{PQ}}\) = P.V of Q – P.V of P
= (α \(\hat{i}\) + β\(\hat{j}\)) – (\(\hat{i}\) + 2 \(\hat{j}\))
= (α – 1) \(\hat{i}\) + (β – 2) \(\hat{j}\)
Also \(\overrightarrow{\mathrm{PQ}}\) = -2 \(\hat{i}\) + 3 \(\hat{j}\)
∴ (α – 1) \(\hat{i}\) + (β – 2) \(\hat{j}\) = -2 \(\hat{i}\) + 3 \(\hat{j}\)
Comparing the coefficients of \(\hat{i}\) and \(\hat{j}\) on both sides, we get
and
α – 1 = -2
⇒ α = -1
β – 2 = 3
⇒ β = 5
Thus the required terminal point be (-1, 5).
Question 10.
If the coordinates of the points A and B in a plane are (1, 1) and (1, 2) respectively, find the coordinates of C (in the plane) such that \(\overrightarrow{A B}\) and \(\overrightarrow{B C}\) are equal.
Answer:
Given P.V of A = \(\hat{i}\) + \(\hat{j}\);
P.V of B = \(\hat{i}\) + 2 \(\hat{j}\)
Let the coordinates of C be (α, β).
∴ P.O of C = α \(\hat{i}\) + β \(\hat{j}\)
Since \(\overrightarrow{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{BC}}\)
⇒ P.B of B – P. of A
= P.V of C – P.V of B
⇒ (\(\hat{i}\) + 2 \(\hat{j}\)) – (\(\hat{i}\) + \(\hat{j}\))
= (α \(\hat{i}\) + β \(\hat{j}\)) – (\(\hat{i}\) + 2 \(\hat{j}\))
⇒ 0 \(\hat{i}\) + \(\hat{j}\)
= (α – 1) \(\hat{i}\) + (β – 2) \(\hat{j}\)
Comparing the coefficients of \(\hat{i}\) and \(\hat{j}\) on both sides, we have
⇒ α-1 = 0 ; β-2 = 1
α = 1 and β = 3
Hence the required coordinates of point C be (1, 3).
Question 11.
If A(1, 1), B(2, 4), C(-1, 2) and D(-1, 3) are the points in a plane, find a point P in the same plane such that \(\overrightarrow{A P}\) = \(\overrightarrow{A B}\) + \(\overrightarrow{C D}\).
Answer:
Question 12.
Prove (by vectors) that the points (-3, 5), (-2, 3) and (4, -9) are collinear.
Answer:
Let the given points be A(-3, 5), B(-2, 3) and C(4, -9) and O be the origin.
Thus \(\overrightarrow{\mathrm{AC}}\)and \(\overrightarrow{\mathrm{AB}}\) are parallel vectors. But A be the common point. Hence A, B and C are collinear.
Question 13.
If\( \vec{a}\) = 7 \(\hat{i}\) – 6 \(\hat{j}\) and \(\vec{b}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\), find ⇒ \(\vec{a}\) + \(\vec{b}\) and determine a unit vector in the direction of the vector \(\vec{a}\) + \(\vec{b}\).
Answer:
Given \(\vec{a}\) = 7 \(\hat{i}\) – 6 \(\hat{j}\) and \(\vec{b}\)
= 3 \(\hat{i}\) + 4 \(\hat{j}\)
∴ \(\vec{a}\) + \(\vec{b}\)
= 7 \(\hat{i}\) – 6 \(\hat{j}\) + 3 \(\hat{i}\) + 4 \(\hat{j}\)
= 10 \(\hat{i}\) – 2 \(\hat{j}\)
⇒ |\(\vec{a}\) + \(\vec{b}\)|
= \(\sqrt{(10)^2+(-2)^2}\)
= \(\sqrt{104}\) = 2 \(\sqrt{26}\)
Thus required unit vector in the direction of \(\vec{a}\) + \(\vec{b}\)
= \(\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}\)
= \(\frac{10 \hat{i}-2 \hat{j}}{2 \sqrt{26}}\)
= \(\frac{5}{\sqrt{26}} \hat{i}-\frac{1}{\sqrt{26}} \hat{j}\)
Question 14.
Prove that the vectors \(\vec{a}\) = -4 \(\hat{i}\) – \(\hat{j}\), \(\vec{b}\) = \(\hat{i}\) – 4 \(\hat{j}\) and \(\vec{c}\) = 3 \(\hat{i}\) + 5 \(\hat{j}\) form a right angled-triangle.
Answer:
Given \(\vec{a}\) = -4 \(\hat{i}\) – \(\hat{j}\) ;
\(\vec{b}\) = \(\hat{i}\) – 4 \(\hat{j}\) and
\(\vec{c}\) = 3 \(\hat{i}\) + 5 \(\hat{j}\)
Here \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)
= -4 \(\hat{i}\) – \(\hat{j}\) + \(\hat{i}\) – 4 \(\hat{j}\) + 3 \(\hat{i}\) + 5 \(\hat{j}\)
= \(\overrightarrow{0}\)
∴ \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are the vectors along the sides of ∆ABC.
a = |\(\vec{a}\)| = \(\sqrt{(-4)^2+(-1)^2}\) = \(\sqrt{17}\) ;
b = |\(\vec{b}\)| = \(\sqrt{1^2+(-4)^2}\) = \(\sqrt{17}\)
and c = |\(\vec{c}\)| = \(\sqrt{3^2+5^2}\) = \(\sqrt{34}\)
∴ a2 + b2 = c2 .
Thus ∆ABC be right angled triangle since pythagoras law holds.
Question 15.
Show that the points
\(\vec{a}\) = -3 \(\sqrt{3}\) \(\hat{i}\) – 3 \(\hat{j}\), \(\vec{b}\) = 6 \(\hat{j}\),
\(\vec{c}\) = 3 \(\sqrt{3}\) \(\hat{i}\) – 3 \(\hat{j}\)
form the sides of an isosceles triangle.
Answer:
Question 16.
Show that the three points with position vectors 2 \(\hat{i}\) + 3 \(\hat{j}\), 3 \(\hat{i}\) + \(\frac{9}{4}\) \(\hat{j}\), 5 \(\hat{i}\) + \(\frac{3}{4}\) \(\hat{j}\) are collinear.
Answer:
