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ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.3
Very short answer type questions (1 to 13) :
Question 1.
(i) If \(\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}\), then find \(|\vec{a} \times \vec{b}|\). (NCERT)
(ii) Find the magnitude of \(\vec{a}\), where \(\vec{a}=(\hat{i}+3 \hat{j}-2 \hat{k}) \times(-\hat{i}+3 \hat{k})\).
(iii) If \(\vec{a}=3 \hat{i}-5 \hat{j}\), \(\vec{b}=6 \hat{i}+3 \hat{j}\) are two vectors and \(\vec{c}\) is a vector such that \(\vec{c}=\vec{a} \times \vec{b}\), then find \(|\vec{a}|:|\vec{b}|:|\vec{c}|\).
Solution:
(i) ∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & 3 \\
3 & 5 & -2
\end{array}\right|\)
= \(\hat{i}(-2-15)-\hat{j}(-4-9)+\hat{k}(10-3)\)
= \(-17 \hat{i}+13 \hat{j}+7 \hat{k}\)
Thus \(|\vec{a} \times \vec{b}|=\sqrt{(-17)^2+(13)^2+7^2}\)
= \(=\sqrt{289+169+49}=\sqrt{507}\)
(ii) \(\vec{a}=(\hat{i}+3 \hat{j}-2 \hat{k}) \times(-\hat{i}+3 \hat{k})\)
∴ \(\vec{a}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 3 & -2 \\
-1 & 0 & 3
\end{array}\right|\)
= \(\hat{i}(9-0)-\hat{j}(3-2)+\hat{k}(0+3)\)
= \(9 \hat{i}-\hat{j}+3 \hat{k}\)
∴ \(|\vec{a}|=\sqrt{9^2+(-1)^2+3^2}\)
= \(\sqrt{81+1+9}=\sqrt{91}\)
(iii) Given \(\vec{a}=3 \hat{i}-5 \hat{j}\) ;
\(\vec{b}=6 \hat{i}+3 \hat{j}\)
∴ \(\vec{c}=\vec{a} \times \vec{b}\)
= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -5 & 0 \\
6 & 3 & 0
\end{array}\right|\)
= \(\hat{i}(0-0)-\hat{j}(0-0)+\hat{k}(9+30)\)
= 39 \(\hat{k}\)
Thus \(|\vec{c}|=|39 \hat{k}|\)
= 39 × 1 = 39
\(|\vec{a}|=\sqrt{3^2+(-5)^2}\)
= \(\sqrt{9+25}=\sqrt{34}\)
\(|\vec{b}|=\sqrt{6^2+3^2}\)
= \(\sqrt{36+9}=\sqrt{45}\)
Thus, \(|\vec{a}|:|\vec{b}|:|\vec{c}|=\sqrt{34}: \sqrt{45}: 39\).
Question 1 (old).
(ii) If \(\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}\) and \(\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}\), then find \(|\vec{a} \times \vec{b}|\).
Solution:
Given \(\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}\)
and \(\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -7 & 7 \\
3 & -2 & 2
\end{array}\right|\)
= \(\hat{i}(0)-\hat{j}(2-21)+\hat{k}(-2+21)\)
= \(19 \hat{j}+19 \hat{k}\)
∴ \(|\vec{a} \times \vec{b}|=\sqrt{(19)^2+(19)^2}=19 \sqrt{2}\)
Question 2.
(i) Find the angle between two vectors \(\vec{a} \text { and } \vec{b}\) with magnitudes 1 and 2 respectively when \(|\vec{a} \times \vec{b}|=\sqrt{3}\).
(ii) If vectors \(\vec{a} \text { and } \vec{b}\) are such that \(|\vec{a}|=3,|\vec{b}|=\frac{2}{3}\) and \(\vec{a} \times \vec{b}\) is a unit vector then find the angle between \(\vec{a} \text { and } \vec{b}\).
(iii) If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 7 and \(\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}\), find the angle between \(\vec{a} \text { and } \vec{b}\).
Solution:
(i) Given \(|\vec{a}|\) = 1 ;
\(|\vec{b}|\) = 2
also, \(\sqrt{3}=|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}|\) sin θ
[∵ \(|\hat{\eta}|\) = 1]
where θ be the angle between \(\vec{a} \text { and } \vec{b}\)
⇒ √3 = 1 × 2 × sin θ
⇒ sin θ = \(\frac{\sqrt{3}}{2}\)
⇒ θ = \(\frac{\pi}{3}\)
(ii) Given \(|\vec{a}|\) =3 ;
\(|\vec{b}|\) = \(\frac{2}{3}\)
and \(\vec{a} \times \vec{b}\) is a unit vector
or \(\vec{a} \times \vec{b}\) = 1
⇒ \(|\vec{a}||\vec{b}|\) sin θ = 1,
where θ be the angle between \(\vec{a} \text { and } \vec{b}\)
⇒ 3 × \(\frac{2}{3}\) sin θ = 1
sin θ = \(\frac{1}{2}\)
θ = \(\frac{\pi}{6}\)
(iii) Given \(|\vec{a}|\) = 2 ;
\(|\vec{b}|\) = 7
and \(\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}\)
= \(\sqrt{9+4+36}\)
= \(\sqrt{49}\) = 7
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\)
Then sin θ = \(\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\)
∴ sin θ = \(\frac{7}{2 \times 7}=\frac{1}{2}\)
θ = 30°, 150°
Question 3.
Find the angle between two vectors \(\vec{a} \text { and } \vec{b}\) if \(|\vec{a} \times \vec{b}|=\vec{a} \cdot \vec{b}\). (NCERT)
Solution:
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\)
Then \(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}|\) sin θ
[∵ \(|\hat{\eta}|\) = 1]
and \(|\vec{a} \cdot \vec{b}|=|\vec{a}||\vec{b}|\) cos θ
Since \(|\vec{a} \times \vec{b}|=|\vec{a} \cdot \vec{b}|\) (given)
⇒ \(|\vec{a}||\vec{b}|\) sin θ = \(|\vec{a}||\vec{b}|\) cos θ
⇒ tan θ = 1
⇒ θ = \(\frac{\pi}{4}\)
Question 4.
(i) Find \(\vec{a} \cdot \vec{b}\) if \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 5 and \(|\vec{a} \times \vec{b}|\) = 8.
(ii) If \(|\vec{a}|\) = 4 ; \(|\vec{b}|\) = 2 and angle between \(|\vec{a}|\) and \(|\vec{b}|\) is \(\frac{\pi}{3}\), find \((\vec{a} \times \vec{b})^2\).
(iii) If vectors \(\vec{a} \text { and } \vec{b}\) are such that \(|\vec{a}|\) = \(\frac{1}{2}\), \(|\vec{b}|=\frac{4}{\sqrt{3}}\) and \(|\vec{a} \times \vec{b}|=\frac{1}{\sqrt{3}}\) then find \(|\vec{a} \cdot \vec{b}|\).
Solution:
(i) Given, \(|\vec{a}|\) = 2,
\(|\vec{b}|\) = 5
and \(|\vec{a} \times \vec{b}|\) = 8
By Lagranges identity, we have
\(|\vec{a} \times \vec{b}|^2=|\vec{a}||\vec{b}|^2-(\vec{a} \cdot \vec{b})^2\)
⇒ 64 = 4 × 25 – \((\vec{a} \cdot \vec{b})^2\)
⇒ \((\vec{a} \cdot \vec{b})^2\) = 36
⇒ \(\vec{a} \cdot \vec{b}\) = 6 (Taking +ve sign)
(ii) Given \(|\vec{a}|\) = 4 ;
\(|\vec{b}|\) = 2
since angle between \(|\vec{a}|\) and \(|\vec{b}|\) be \(\frac{\pi}{3}\)
(iii) Given \(|\vec{a}|\) = \(\frac{1}{2}\) ;
\(|\vec{b}|=\frac{4}{\sqrt{3}}\)
and \(|\vec{a} \times \vec{b}|=\frac{1}{\sqrt{3}}\)
Question 4 (old).
(ii) If \(|\vec{a}|\) = 5, \(|\vec{b}|\) = 13 and \(|\vec{a} \times \vec{b}|\) = 25, then find \(\vec{a} \cdot \vec{b}\).
Solution:
Given \(|\vec{a}|\) = 5,
\(|\vec{b}|\) = 13
and \(|\vec{a} \times \vec{b}|\) = 25
If θ be the angle between \(\vec{a} \text { and } \vec{b}\)
sin θ = \(\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\)
= \(\frac{25}{5 \times 13}=\frac{5}{13}\)
∴ cos θ = \(\sqrt{1-\frac{25}{169}}=\frac{12}{13}\)
∴ \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|\) cos θ
= 5 × 13 × \(\frac{12}{13}\) = 60.
Question 5.
(i) Find the value of λ if \((2 \hat{i}+6 \hat{j}+14 \hat{k}) \times(\hat{i}-\lambda \hat{j}+7 \hat{k})=\overrightarrow{0}\).
(ii) Find the value of p if \((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+3 \hat{j}+p \hat{k})=\overrightarrow{0}\).
(iii) Find λ and µ \((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}\). (NCERT)
Solution:
(i) Let \(\vec{a}=2 \hat{i}+6 \hat{j}+14 \hat{k}\) ;
\(\vec{b}=(\hat{i}-\lambda \hat{j}+7 \hat{k})\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 14 \\
1 & -\lambda & 7
\end{array}\right|\)
= \(\hat{i}(42+14 \lambda)-\hat{j}(14-14)+\hat{k}(-2 \lambda-6)\)
= \(\hat{i}(42+14 \lambda)+0 \hat{j}+(-2 \lambda-6) \hat{k}\)
It is given that \(\vec{a} \times \vec{b}=\overrightarrow{0}\)
⇒ \((42+14 \lambda) \hat{i}+0 \hat{j}+(-2 \lambda-6) \hat{k}\) = \(0 \hat{i}+0 \hat{j}+0 \hat{k}\)
∴ 42 + 14λ = 0
⇒ λ = – 3
and – 2λ – 6 = 0
λ = 3
(ii) Let \(\vec{a}=2 \hat{i}+6 \hat{j}+27 \hat{k}\) ;
\(\vec{b}=\hat{i}+3 \hat{j}+p \hat{k}\)
(iii) Let \(\vec{a}=2 \hat{i}+6 \hat{j}+27 \hat{k}\)
and \(\vec{b}=\hat{i}+\lambda \hat{j}+\mu \hat{k}\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 27 \\
1 & \lambda & \mu
\end{array}\right|\)
= \(\hat{i}(6 \mu-27 \lambda)-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)\)
also \(\vec{a} \times \vec{b}=\overrightarrow{0}\)
∴ \((6 \mu-27 \lambda) \hat{i}-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)=\overrightarrow{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}\) ……………….(1)
Thus 6µ – 27λ = 0 ………………………(1)
– (2µ – 27) = 0
⇒ µ = \(\frac{27}{2}\)
and 2λ – 6 = 0
⇒ λ = 3
Also, λ = 3, µ = \(\frac{27}{2}\) satisfies eqn. (1).
Question 6.
(i) Find a unit vector perpendicular to the two vectors \(\hat{i}+2 \hat{j}-\hat{k}\) and \(2 \hat{i}+3 \hat{j}+\hat{k}\). (ISC 2004)
(ii) Find a unit vector perpendicular to \(\vec{a} \text { and } \vec{b}\), where \(\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}\) and \(\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}\).
Solution:
(i) Let \(\vec{a}=\hat{i}+2 \hat{j}-\hat{k}\) ;
\(\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}\)
(ii) Given \(\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}\)
and \(\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -7 & 7 \\
3 & -2 & 2
\end{array}\right|\)
= \(\hat{i}(-14+14)-\hat{j}(2-21)+\hat{k}(-2+21)\)
= \(0 \hat{i}+19 \hat{j}+19 \hat{k}\)
Thus unit vector = \(\pm \frac{\vec{a} \times \vec{b}}{|\vec{a}+\vec{b}|}\)
= \(\pm \frac{19(\hat{j}+\hat{k})}{\sqrt{361+361}}\)
= \(\pm \frac{(\hat{j}+\hat{k})}{\sqrt{2}}\)
(ii) Find a unit vector perpendicular to the plane of \(\vec{a} \text { and } \vec{b}\), where \(\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}\) and \(\vec{b}=\hat{i}-3 \hat{j}+\hat{k}\).
Solution:
Given latex]\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}[/latex]
and \(\vec{b}=\hat{i}-3 \hat{j}+\hat{k}\)
Question 7.
(i) What conclusion can you draw about vectors \(\vec{a} \text { and } \vec{b}\) when \(\vec{a} \times \vec{b}=\overrightarrow{0}\) and \(\vec{a} \cdot \vec{b}\) = 0? (NCERT)
(ii) If either \(\vec{a}\) = 0 or \(\vec{b}\) = 0, then \(\vec{a} \times \vec{b}=\overrightarrow{0}\). Is the converse true? Justify your answer with an example. (NCERT)
Solution:
(i) When \(\vec{a} \times \vec{b}=\overrightarrow{0}\)
⇒ \(\vec{a}\) = 0
or \(\vec{a} \| \vec{b}\)
or \(\vec{a} \cdot \vec{b}\) = 0
⇒ \(\vec{a}=\overrightarrow{0}\)
or \(\vec{b}=\overrightarrow{0}\)
or \(\vec{a} \perp \vec{b}\)
Now a vector can never be parallel and perpendicular simultaneously.
Thus \(\vec{a}=\overrightarrow{0}\) or \(\vec{b}=\overrightarrow{0}\) = 0.
(ii) Given \(\vec{a}=\overrightarrow{0}\)
or \(\vec{b}=\overrightarrow{0}\)
∴ \(\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}\) = 0
Let \(\vec{a}=\hat{i}\) ;
b = 2 \(\hat{i}\)
Here \(\)
but neither \(\vec{a}=\overrightarrow{0}\) nor \(\vec{b}=\overrightarrow{0}\).
Question 7 (old).
(i) show that \(\vec{a} \times \vec{b}=\vec{a} \times \vec{c}\) may not imply that \(\vec{b}=\vec{c}\).
(ii) If \(\vec{b}=\vec{c}\), is \(\vec{a} \times \vec{b}=\vec{c} \times \vec{a}\) ?
Solution:
(i)
(ii) Not necessary ;
Question 8.
Find λ such that \(\vec{a}=\hat{i}+\lambda \hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+2 \hat{j}+9 \hat{k}\) are parallel.
Solution:
Given \(\vec{a}=\hat{i}+\lambda \hat{j}+3 \hat{k}\)
and \(\vec{b}=3 \hat{i}+2 \hat{j}+9 \hat{k}\)
Since \(\vec{a} \text { and } \vec{b}\) are parallel.
∴ \(\vec{a} \times \vec{b}=\overrightarrow{0}\) …………….(1)
Here, \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & \lambda & 3 \\
3 & 2 & 9
\end{array}\right|\)
= \(\hat{i}(9 \lambda-6)-\hat{j}(9-9)+\hat{k}(2-3 \lambda)\)
= \((9 \lambda-6) \hat{i}+(2-3 \lambda) \hat{k}\)
∴ from (1) ; we have
\((9 \lambda-6) \hat{i}+(2-3 \lambda) \hat{k}\)
= \(\overrightarrow{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}\)
∴ 9λ – 6 = 0
⇒ λ = \(\frac{2}{3}\)
and 2 – 3λ = 0
⇒ λ = \(\frac{2}{3}\)
Question 9.
(i) \((\hat{i} \times \hat{j}) \cdot \hat{k}+\hat{i} \cdot \hat{j}\)
(ii) \((\hat{j} \times \hat{k}) \cdot \hat{i}+(\hat{i} \times \hat{k}) \cdot \hat{j}+(\hat{i} \times \hat{j}) \cdot \hat{k}\)
Solution:
Since \(\hat{i}, \hat{j}, \hat{k}\) forms right handed orthogonal system
∴ \(\hat{i} \times \hat{j}=\hat{k}\)
\(\hat{j} \times \hat{k}=\hat{i}\)
\(\hat{k} \times \hat{i}=\hat{j}\)
and \(\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}\) = 0
(i) \((\hat{i} \times \hat{j}) \cdot \hat{k}+\hat{i} \cdot \hat{j}\)
\(\hat{k} \cdot \hat{k}+\hat{i} \cdot \hat{j}\)
= 1 + 0 = 1
[∵ \(\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}\) = 1
and \(\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}\) = 0]
(ii) \((\hat{j} \times \hat{k}) \cdot \hat{i}+(\hat{i} \times \hat{k}) \cdot \hat{j}+(\hat{i} \times \hat{j}) \cdot \hat{k}\)
\(\hat{i} \cdot \hat{i}+(-\hat{j}) \cdot \hat{j}+(\hat{k} \cdot \hat{k})\)
= 1 – 1 + 1 = 0
Question 10.
Find the area of the parallelogram whose adjacent sides are represented by the vectors
(i) \(3 \hat{i}+\hat{j}+4 \hat{k} \text { and } \hat{i}-\hat{j}+\hat{k}\) (NCERT)
(ii) \(2 \hat{i}-3 \hat{k} \text { and } 4 \hat{j}+2 \hat{k}\)
Solution:
(i) Let \(\vec{a}=3 \hat{i}+\hat{j}+4 \hat{k}\)
and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\)
(ii) Let \(\vec{a}=2 \hat{i}-3 \hat{k}\)
and \(\vec{b}=4 \hat{j}+2 \hat{k}\)
Question 11.
Find the area of a triangle whose two sides are represented by the vectors \(3 \hat{i}+\hat{j}+4 \hat{k}\) and \(\hat{i}-\hat{j}+\hat{k}\).
Solution:
Let \(\vec{a}=3 \hat{i}+\hat{j}+4 \hat{k}\)
and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & 4 \\
1 & -1 & 1
\end{array}\right|\)
= \(\hat{i}(1+4)-\hat{j}(3-4)+\hat{k}(-3-1)\)
∴ \(\vec{a} \times \vec{b}=5 \hat{i}+\hat{j}-4 \hat{k}\)
Thus, \(|\vec{a} \times \vec{b}|=\sqrt{5^2+1^2+(-4)^2}\)
= \(\sqrt{25+1+16}=\sqrt{42}\)
Therefore, area of triangle = \(\frac{1}{2}|\vec{a} \times \vec{b}|\)
= \(\frac{1}{2} \sqrt{42}\) sq. units
Question 12.
(i) If \(\vec{a}=4 \hat{i}+3 \hat{j}+2 \hat{k}\) and \(\vec{b}=3 \hat{i}+2 \hat{k}\), then find \(|\vec{b} \times 2 \vec{a}|\).
(ii) If \(\vec{a}=\hat{i}+\hat{j}-3 \hat{k}\) and \(\vec{b}=\hat{j}+2 \hat{k}\), then find \(|2 \vec{b} \times \vec{a}|\).
Solution:
(i) Given, \(\vec{a}=4 \hat{i}+3 \hat{j}+2 \hat{k}\)
and \(\vec{b}=3 \hat{i}+2 \hat{k}\)
(ii) Given, \(\vec{a}=\hat{i}+\hat{j}-3 \hat{k}\) ;
\(\vec{b}=\hat{j}+2 \hat{k}\)
∴ \(2 \vec{b}=2 \hat{j}+4 \hat{k}\)
\(2 \vec{b} \times \vec{a}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & 2 & 4 \\
1 & 1 & -3
\end{array}\right|\)
= \(\hat{i}(-6-4)-\hat{j}(-4)+\hat{k}(-2)\)
= \(-10 \hat{i}+4 \hat{j}-2 \hat{k}\)
∴ \(|2 \vec{b} \times \vec{a}|=\sqrt{(-10)^2+(4)^2+(-2)^2}\)
= \(\sqrt{100+16+4}\)
= \(\sqrt{120}=2 \sqrt{30}\)
Question 13.
(i) If \(\vec{a}=3 \hat{i}-\hat{j}-2 \hat{k}\) and \(\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}\), then find \((\vec{a}+2 \vec{b}) \times(2 \vec{a}-\vec{b})\).
(ii) Taking \(\vec{a}=2 \hat{i}-3 \hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}+4 \hat{j}-2 \hat{k}\) verify that \(\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}\).
Solution:
(i) Given \(\vec{a}=3 \hat{i}-\hat{j}-2 \hat{k}\) ;
\(\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}\)
\(\vec{a}+2 \vec{b}=(3 \hat{i}-\hat{j}-2 \hat{k})+2(2 \hat{i}+3 \hat{j}+\hat{k})\)
= \(7 \hat{i}+5 \hat{j}+0 \hat{k}\)
\(2 \vec{a}-\vec{b}=2(3 \hat{i}-\hat{j}-2 \hat{k})-(2 \hat{i}+3 \hat{j}+\hat{k})\)
\(4 \hat{i}-5 \hat{j}-5 \hat{k}\)
Thus \((\vec{a}+2 \vec{b}) \times(2 \vec{a}-\vec{b})\)
= \(\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
7 & 5 & 0 \\
4 & -5 & -5
\end{array}\right|\)
= \(\hat{i}(-25-0)-\hat{j}(-35-0)+\hat{k}(-35-20)\)
= \(-25 \hat{i}+35 \hat{j}-55 \hat{k}\)
(ii) Given \(\vec{a}=2 \hat{i}-3 \hat{j}-\hat{k}\)
and \(\vec{b}=\hat{i}+4 \hat{j}-2 \hat{k}\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -3 & -1 \\
1 & 4 & -2
\end{array}\right|\)
= \(\hat{i}(6+4)-\hat{j}(-4+1)+\hat{k}(8+3)\)
= \(10 \hat{i}+3 \hat{j}+11 \hat{k}\)
∴ \(\vec{b} \times \vec{a}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 4 & -2 \\
2 & -3 & -1
\end{array}\right|\)
= \(\hat{i}(-4-6)-\hat{j}(-1+4)+\hat{k}(-3-8)\)
= \(-10 \hat{i}-3 \hat{j}-11 \hat{k}\)
∴ \(\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}\)
Question 14.
If θ is the angle between two vectors \(\hat{i}-2 \hat{j}+3 \hat{k}\) and \(3 \hat{i}-2 \hat{j}+\hat{k}\), find sin θ.
Solution:
Let \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\)
and \(\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k}\)
Question 15.
If \(\vec{a}=3 \hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{b}=2 \hat{i}-2 \hat{j}+4 \hat{k}\), then find :
(i) the sine of the angle between \(\vec{a} \text { and } \vec{b}\). (NCERT Exemplar)
(ii) a unit vector perpendicular to both the vectors \(\vec{a} \text { and } \vec{b}\).
Solution:
(i) Given \(\vec{a}=3 \hat{i}+\hat{j}+2 \hat{k}\)
and \(\vec{b}=2 \hat{i}-2 \hat{j}+4 \hat{k}\)
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\)
(ii) Here,
\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & 2 \\
2 & -2 & 4
\end{array}\right|\)
= \(8 i-8 j-8 k\)
\(|\vec{a} \times \vec{b}|=\sqrt{8^2+(-8)^2+(-8)^2}\)
= 8√3
Thus required unit vector ⊥to both \(\vec{a} \text { and } \vec{b}\)
= \(\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \pm \frac{8(\hat{i}-\hat{j}-\hat{k})}{8 \sqrt{3}}\)
= ± \(\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})\)
Question 15 (old).
(i) If \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\), \(\vec{b}=\hat{i}+2 \hat{j}-\hat{k}\) and \(\vec{c}=2 \hat{i}+3 \hat{j}\), find \((\vec{a} \times \vec{b}) \times \vec{c}\) and \(\vec{a} \times(\vec{b} \times \vec{c})\), and verify that these are not the same.
(ii) If \(\vec{a}, \vec{b} \text { and } \vec{c}\) represent the position vectors of the points with coordinates (2, – 10, 2), (3, 1, 2) and (2, 1, 3) respectively, find the value of \(\vec{a} \times(\vec{b} \times \vec{c})\). (ISC 2009)
Solution:
(i) Given \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\),
\(\vec{b}=\hat{i}+2 \hat{j}-\hat{k}\)
and \(\vec{c}=2 \hat{i}+3 \hat{j}\)
\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
1 & 2 & -1
\end{array}\right|\)
= \(\hat{i}(1-2)-\hat{j}(-2-1)+\hat{k}(4+1)\)
= \(-\hat{i}+3 \hat{j}+5 \hat{k}\)
(ii) Given \(\vec{a}=2 \hat{i}-10 \hat{j}+2 \hat{k}\),
\(\vec{b}=3 \hat{i}+\hat{j}+2 \hat{k}\)
and \(\vec{c}=2 \hat{i}+\hat{j}+3 \hat{k}\)
∴ \(\vec{b} \times \vec{c}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & 2 \\
2 & 1 & 3
\end{array}\right|\)
= \(\hat{i}(3-2)-\hat{j}(9-4)+\hat{k}(3-2)\)
= \(\hat{i}-5 \hat{j}+\hat{k}\)
∴ \(\vec{a} \times(\vec{b} \times \vec{c})=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & -10 & 2 \\
1 & -5 & 1
\end{array}\right|\)
= \(\hat{i}(-10+10)-\hat{j}(2-2)+\hat{k}(-10+10)\)
= \(0 \hat{i}+0 \hat{j}+0 \hat{k}\)
= \(\overrightarrow{0}\)
Question 16.
(i) Find a vector of magnitude 6 units which is perpendicular to both the vectors \(2 \hat{i}-\hat{j}+2 \hat{k}\) and \(4 \hat{i}-\hat{j}+3 \hat{k}\) (NCERT Exemplar)
(ii) Find a vector of magnitude 19 which is perpendicular to both the vectors \(-\hat{j}+\hat{k}\) and \(4 \hat{i}-\hat{j}+8 \hat{k}\).
Solution:
(i) Let \(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}\)
and \(\vec{b}=4 \hat{i}-\hat{j}+3 \hat{k}\)
(ii) Let \(\vec{a}=4 \hat{i}-\hat{j}+8 \hat{k}\)
and \(\vec{b}=-\hat{j}+\hat{k}\)
Question 17.
Find a vector of magnitude 6, perpendicular to each of the vectors \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\) where \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\).
Solution:
Given, \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\)
and \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\)
Question 18.
If \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k}\), find \(\vec{a} \times \vec{b}\). Verify that \(\vec{a}\) and \(\vec{a} \times \vec{b}\) are perpendicular to each other.
Solution:
Given \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\)
and \(\vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k}\)
\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 3 \\
2 & 3 & -5
\end{array}\right|\)
= \(\hat{i}(10-9)-\hat{j}(-5-6)+\hat{k}(3+4)\)
= \(\hat{i}+11 \hat{j}+7 \hat{k}\)
Now \(\vec{a} \cdot(\vec{a} \times \vec{b})\)
= \((\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+11 \hat{j}+7 \hat{k})\)
= 1 (1) – 2 (11) + 3 (7) = 0
∴ \(\vec{a} \perp \vec{a} \times \vec{b}\)
Question 18 (old).
Find a unit vector perpendicular to the vectors \(4 \hat{i}+3 \hat{j}+\hat{k}\) and \(2 \hat{i}-\hat{j}+2 \hat{k}\). Determine the sine of the angle between these two vectors. (ISC 2007)
Solution:
Given \(\vec{a}=4 \hat{i}+3 \hat{j}+\hat{k}\)
and \(\vec{b}=2 \hat{i}-\hat{j}+2 \hat{k}\)
Question 19.
(i) If \(\vec{a}=\hat{i}-\hat{j}\), \(\vec{b}=3 \hat{j}-\hat{k}\) and \(\vec{c}=7 \hat{i}-\hat{k}\), find a vector \(\vec{d}\) which is perpendicular to both \(\vec{a} \text { and } \vec{b}\) and \(\vec{c} \cdot \vec{d}\) = 1.
(ii) If \(\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}\), \(\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}\), and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\), find a vector \(\vec{d}\) which is perpendicular to both the vectors \(\vec{a} \text { and } \vec{b}\) and \(\vec{c} \cdot \vec{d}\) = 27.
(iii) If \(\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}\), \(\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}\) and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\). Find a vector \(\vec{d\) which is perpendicular to both \(\vec{a} \text { and } \vec{b}\) and \(\vec{c} \cdot \vec{d}\) = 18.
(iv) If \(\vec{a}=4 \hat{i}+5 \hat{j}-\hat{k}\), \(\vec{b}=\hat{i}-4 \hat{j}+5 \hat{k}\) and \(\vec{c}=3 \hat{i}+\hat{j}-\hat{k}\). Find a vector \(\vec{d\) which is perpendicular to both \(\vec{c} \text { and } \vec{b}\) and \(\vec{d} \cdot \vec{a}\) = 21.
Solution:
(i) Given \(\vec{a}=\hat{i}-\hat{j}\),
\(\vec{b}=3 \hat{j}-\hat{k}\)
and \(\vec{c}=7 \hat{i}-\hat{k}\)
(ii) Given \(\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}\),
\(\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}\),
and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\)
Let the required vector be \(\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}\)
such that \(\vec{d}\) is perpendicular to \(\vec{a} \text { and } \vec{b}\)
∴ \(\vec{d} \cdot \vec{b}\) = 0
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+4 \hat{j}+2 \hat{k})\) = 0
⇒ x + 4y + 7z = 0 …………………(1)
Also \(\vec{d} \cdot \vec{b}\) = 0
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+7 \hat{k})\) = 0
⇒ 3x – 2y + 7z = 0 ………………..(2)
also given \(\vec{c} \cdot \vec{d}\) = 27
⇒ \((2 \hat{i}-\hat{j}+4 \hat{k}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})\)
⇒ 2x – y + 4z = 0 ………………..(3)
Multiplying eqn. (2) by 2 and adding to eqn. (1) ; we get
7x + 16z = 0 …………………..(4)
Multiplying eqn. (3) by 2 ; we have
4x – 2y + 8z = 54 …………………….(5)
eqn. (5) – eqn. (2) gives ;
x + z = 54 ……………….(6)
Multiplying eqn. (6) by (7) – eqn. (4) ; gives
⇒ 7z – 16z = + 54 × 7
⇒ – 9z = + 54 × 7
⇒ z = – 42
∴ from eqn. (6) ;
x = 96
∴ from eqn. (3) ;
y = 192 – 168 – 27 = – 3
Thus required vector be \(\vec{d}=96 \hat{i}-3 \hat{j}-42 \hat{k}\)
i.e., \(\vec{d}=3(32 \hat{i}-\hat{j}-14 \hat{k})\)
(iii) Given \(\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}\),
\(\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}\)
and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\)
Let the required vector be
\(\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}\) …………………..(1)
Since \(\vec{d}\) is perpendicular to \(\vec{a} \text { and } \vec{b}\)
∴ \(\vec{d} \cdot \vec{b}\) = 0
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+4 \hat{j}+2 \hat{k})\) = 0
⇒ x + 4y + 2z = 0 ………………….(2)
Also \(\vec{d} \cdot \vec{b}\) = 0
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+7 \hat{k})\) = 0
⇒ 3x – 2y + 7z = 0 ……………………..(3)
given \(\vec{c} \cdot \vec{d}\) = 18
⇒ \((2 \hat{i}-\hat{j}+4 \hat{k}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})\) = 18
⇒ 2x – y + 4z = 18 ……………………..(4)
Multiplying eqn. (3) by (2) and adding to eqn. (2) ; we have
7x + 16z = 0 ……………………(5)
Multiplying eqn. (4) by (2) – eqn. (3) ; we have
x + z = 36 ……………………..(6)
Multiply eqn. (6) by 7 – eqn. (5) ; we have
– 9z = 36 × 7
⇒ z = – 28
∴ from (6) ;
x = 64
∴ from (4) ;
y = 128 – 112 – 18 = – 2
∴ from (1) ;
\(\vec{d}=64 \hat{i}-2 \hat{j}-28 \hat{k}\)
= \(2(32 \hat{i}-\hat{j}-14 \hat{k})\)
(iv) Given \(\vec{a}=4 \hat{i}+5 \hat{j}-\hat{k}\),
\(\vec{b}=\hat{i}-4 \hat{j}+5 \hat{k}\)
and \(\vec{c}=3 \hat{i}+\hat{j}-\hat{k}\)
Let \(\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}\)
Since \(\vec{d}\) is perpendicular to \(\vec{c}\)
⇒ \(\vec{d} \text { and } \vec{c}\) = 0
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}+\hat{j}-\hat{k})\) = 0
⇒ 3x + y – z = 0 ………………..(1)
Also \(\vec{d}\) is perpendicular to \(\vec{b\)
⇒ \(\vec{d} \text { and } \vec{b}\) = 0
i.e. \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}-4 \hat{j}+5 \hat{k})\) = 0
⇒ x – 4y + 5z = 0 …………………..(2)
Also, \(\) = 21
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(4 \hat{i}+5 \hat{j}-\hat{k})\) = 21
⇒ 4x + 5y – z = 21 ……………………(3)
Multiply eqn. (1) by 5 and adding to eqn. (2) ; we get
16x + y = 0 ………………….(4)
Multiply eqn. (3) by 5 and adding to eqn. (2) ; we get
21x + 21y = 105
⇒ x + y = 5 ……………………(5)
On solving eqn. (4) and (5) ; we have
x = – \(\frac{1}{3}\)
and y = \(\frac{16}{3}\)
Question 20.
Define \(\vec{a} \times \vec{b}\) and prove that \(|\vec{a} \times \vec{b}|=(\vec{a} \cdot \vec{b})\) tan θ, where θ is the angle between \(\vec{a} \text { and } \vec{b}\).
Solution:
The vector product \(\) is a vector quantity whose magnitude is \(|\vec{a}||\vec{b}|\) sin θ where θ is the angle between \(\vec{a} \text { and } \vec{b}\) and whose direction is perpendicular to plane of \(\vec{a} \text { and } \vec{b}\) in such a way that \(\vec{a}, \vec{b}\) and this direction forms a right handed system.
∴ \(\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}\)
where \(\hat{n\) be the unit vector ⊥ to the plane of \(\vec{a} \text { and } \vec{b}\) s.t. \(\vec{a}, \vec{b}, \hat{n}\) forms a right handed system.
Since \(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}|\) ……………………..(1)
We know that \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)
∴ \(|\vec{a}||\vec{b}|=\frac{\vec{a} \cdot \vec{b}}{\cos \theta}\)
∴ from (1) ;
\(|\vec{a} \times \vec{b}|=\frac{\vec{a} \cdot \vec{b}}{\cos \theta}\) sin θ
∴ \(|\vec{a} \times \vec{b}|=(\vec{a} \cdot \vec{b})\) tan θ
Question 21.
In ∆ OAB, \(\overrightarrow{\mathrm{OA}}=3 \hat{i}+2 \hat{j}-\hat{k}\) and \(\overrightarrow{\mathrm{OB}}=\hat{i}+3 \hat{j}+\hat{k}\). Find the area of the triangle OAB.
Solution:
Given \(\overrightarrow{\mathrm{OA}}=3 \hat{i}+2 \hat{j}-\hat{k}\)
and \(\overrightarrow{\mathrm{OB}}=\hat{i}+3 \hat{j}+\hat{k}\)
Question 22.
Using vectors, find the area of the triangle whose vertices are : A (3, – 1, 2), B (1, – 1, – 3), C (4, – 3, 1). (ISC 2020)
Solution:
Question 23.
(i) Find the area of a parallelogram whose diagonals are determined by the vectors \(\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}\) and \(\vec{b}=\hat{i}-3 \hat{j}+4 \hat{k}\). (ISC 2009)
(ii) The vectors \(-2 \hat{i}+4 \hat{j}+4 \hat{k}\) and \(-4 \hat{i}-2 \hat{k}\) represent the diagonals BD and AC of a parallelogram ABCD. Find the area of the parallelogram. (ISC 2006)
Solution:
(i) Given \(\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}\)
and \(\vec{b}=\hat{i}-3 \hat{j}+4 \hat{k}\)
Now \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & -2 \\
1 & -3 & 4
\end{array}\right|\)
= \(\hat{i}(4-6)-\hat{j}(12+2)+\hat{k}(-9-1)\)
= \(-2 \hat{i}-14 \hat{j}-10 \hat{k}\)
∴ Required area of ||gm = \(\frac{1}{2}|\vec{a} \times \vec{b}|\)
= \(\frac{1}{2} \sqrt{4+196+100}=5 \sqrt{3}\)
(ii) Given \(\vec{a}=-2 \hat{i}+4 \hat{j}+4 \hat{k}\)
and \(\vec{b}=-4 \hat{i}-2 \hat{k}\)
Now, \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-2 & 4 & 4 \\
-4 & 0 & -2
\end{array}\right|\)
= \(\hat{i}(-8)-\hat{j}(4+16)+\hat{k}(16)\)
= \(-8 \hat{i}-20 \hat{j}+16 \hat{k}\)
= \(4(-2 \hat{i}-5 \hat{j}+4 \hat{k})\)
∴ Required area of ||gm = \(\frac{1}{2}|\vec{a} \times \vec{b}|\)
= \(\frac{\sqrt{64+400+256}}{2}\)
= \(\frac{\sqrt{720}}{2}\)
= \(\frac{12 \sqrt{5}}{2}\)
= 6√5
Question 24.
The two adjacent sides of a parallelogram are represented by the vectors \(2 \hat{i}-4 \hat{j}-5 \hat{k}\) and \(2 \hat{i}+2 \hat{j}+3 \hat{k}\).
Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.
Solution:
Given, \(\vec{a}=2 \hat{i}-4 \hat{j}-5 \hat{k}\)
and \(\vec{b}=2 \hat{i}+2 \hat{j}+3 \hat{k}\)
Question 25.
Find the area of the triangle formed by the points A (1, 2, 3), B(2, -1 , 4) and C (4, 5, – 1).
Solution:
Given P.V. of A = \(\hat{i}+2 \hat{j}+3 \hat{k}\) ;
P.V. of B = \(2 \hat{i}-\hat{j}+4 \hat{k}\)
and P.V. of C = \(4 \hat{i}+5 \hat{j}-\hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \((2 \hat{i}-\hat{j}+4 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})\)
= \(\hat{i}-3 \hat{j}+\hat{k}\)
\(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \((4 \hat{i}+5 \hat{j}-\hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})\)
= \(3 \hat{i}+3 \hat{j}-4 \hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -3 & 1 \\
3 & 3 & -4
\end{array}\right|\)
= \(\hat{i}(12-3)-\hat{j}(-4-3)+\hat{k}(3+9)\)
= \(9 \hat{i}+7 \hat{j}+12 \hat{k}\)
∴ \(|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{9^2+7^2+12^2}=\sqrt{274}\)
Thus required area of ∆ABC = \(\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|\)
= \(\frac{\sqrt{274}}{2}\)