Class 12 Ch 10: Vector Algebra - Exercise 10.4

Q1

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Find $|\vec{a} \times \vec{b}|$, if $\vec{a} = i - 7j + 7k$ and $\vec{b} = 3i - 2j + 2k$.

$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix}$

$= i(-14 - (-14)) - j(2 - 21) + k(-2 - (-21))$

$= 0i + 19j + 19k$.

$|\vec{a} \times \vec{b}| = \sqrt{0^2 + 19^2 + 19^2} = \sqrt{2(19^2)} = 19\sqrt{2}$.

$\boxed{19\sqrt{2}}$

Q2

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Find a unit vector perpendicular to each of the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$, where $\vec{a} = 3i + 2j + 2k$ and $\vec{b} = i + 2j - 2k$.

Let $\vec{p} = \vec{a} + \vec{b} = 4i + 4j$ and $\vec{q} = \vec{a} - \vec{b} = 2i + 4k$.

$\vec{p} \times \vec{q} = \begin{vmatrix} i & j & k \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = i(16) - j(16) + k(-8) = 16i - 16j - 8k$.

$|\vec{p} \times \vec{q}| = \sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576} = 24$.

Unit vector = $\pm \frac{16i - 16j - 8k}{24} = \pm (\frac{2}{3}i - \frac{2}{3}j - \frac{1}{3}k)$.

$\boxed{\pm (\frac{2}{3}i - \frac{2}{3}j - \frac{1}{3}k)}$

Q3

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If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $i$, $\frac{\pi}{4}$ with $j$ and an acute angle $\theta$ with $k$, then find $\theta$ and hence, the components of $\vec{a}$.

$l = \cos \frac{\pi}{3} = \frac{1}{2}$, $m = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, $n = \cos \theta$.

$l^2 + m^2 + n^2 = 1 \Rightarrow \frac{1}{4} + \frac{1}{2} + n^2 = 1 \Rightarrow n^2 = \frac{1}{4}$.

$n = \pm \frac{1}{2}$. Since $\theta$ is acute, $n = \frac{1}{2}$, so $\theta = \frac{\pi}{3}$.

Components are $(l, m, n)$.

$\boxed{\theta = \frac{\pi}{3}; \text{ Components: } (\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2})}$

Q4

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Show that $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b})$.

LHS $= \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}$.

Since $\vec{a} \times \vec{a} = \vec{0}$ and $\vec{b} \times \vec{b} = \vec{0}$.

$= \vec{0} + \vec{a} \times \vec{b} - (-\vec{a} \times \vec{b}) - \vec{0} = 2(\vec{a} \times \vec{b})$.

$\boxed{\text{Shown}}$

Q5

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Find $\lambda$ and $\mu$ if $(2i + 6j + 27k) \times (i + \lambda j + \mu k) = \vec{0}$.

$\begin{vmatrix} i & j & k \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} = i(6\mu - 27\lambda) - j(2\mu - 27) + k(2\lambda - 6) = \vec{0}$.

$2\lambda - 6 = 0 \Rightarrow \lambda = 3$.

$2\mu - 27 = 0 \Rightarrow \mu = \frac{27}{2}$.

$\boxed{\lambda = 3, \mu = \frac{27}{2}}$

Q6

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Given that $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \times \vec{b} = \vec{0}$. What can you conclude about the vectors $\vec{a}$ and $\vec{b}$?

$\vec{a} \cdot \vec{b} = 0 \Rightarrow \vec{a} \perp \vec{b}$ or one is zero.

$\vec{a} \times \vec{b} = \vec{0} \Rightarrow \vec{a} \parallel \vec{b}$ or one is zero.

Vectors cannot be both perpendicular and parallel simultaneously unless one of them is the zero vector.

$\boxed{\text{Either } \vec{a} = \vec{0} \text{ or } \vec{b} = \vec{0}}$

Q7

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Let the vectors $\vec{a}, \vec{b}, \vec{c}$ be given as $a_1i + a_2j + a_3k$, $b_1i + b_2j + b_3k$, $c_1i + c_2j + c_3k$. Then show that $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$.

Using properties of determinants: $\vec{a} \times (\vec{b} + \vec{c}) = \begin{vmatrix} i & j & k \\ a_1 & a_2 & a_3 \\ b_1+c_1 & b_2+c_2 & b_3+c_3 \end{vmatrix}$.

Splitting the determinant along the third row: $= \begin{vmatrix} i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} + \begin{vmatrix} i & j & k \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$.

$= \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$.

$\boxed{\text{Shown}}$

Q8

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If either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then $\vec{a} \times \vec{b} = \vec{0}$. Is the converse true? Justify your answer with an example.

The converse is not true. $\vec{a} \times \vec{b} = \vec{0}$ implies $\vec{a} \parallel \vec{b}$ (collinear), not necessarily that one is zero.

Example: $\vec{a} = 2i + 2j + 2k$, $\vec{b} = i + j + k$.

$\vec{a} \times \vec{b} = \vec{0}$, but $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$.

$\boxed{\text{No}}$

Q9

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Find the area of the triangle with vertices $A(1, 1, 2)$, $B(2, 3, 5)$ and $C(1, 5, 5)$.

$\vec{AB} = i + 2j + 3k$, $\vec{AC} = 0i + 4j + 3k$.

$\vec{AB} \times \vec{AC} = \begin{vmatrix} i & j & k \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} = i(6-12) - j(3-0) + k(4-0) = -6i - 3j + 4k$.

Area $= \frac{1}{2}|\vec{AB} \times \vec{AC}| = \frac{1}{2}\sqrt{36 + 9 + 16} = \frac{\sqrt{61}}{2}$.

$\boxed{\frac{\sqrt{61}}{2} \text{ sq. units}}$

Q10

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Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a} = i - j + 3k$ and $\vec{b} = 2i - 7j + k$.

Area $= |\vec{a} \times \vec{b}|$.

$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} = i(20) - j(-5) + k(-5) = 20i + 5j - 5k$.

Magnitude $= \sqrt{400 + 25 + 25} = \sqrt{450} = 15\sqrt{2}$.

$\boxed{15\sqrt{2} \text{ sq. units}}$

Q11

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Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}| = 3$ and $|\vec{b}| = \frac{\sqrt{2}}{3}$, then $\vec{a} \times \vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is
(A) $\pi/6$
(B) $\pi/4$
(C) $\pi/3$
(D) $\pi/2$

$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin \theta = 1$.

$3 \times \frac{\sqrt{2}}{3} \sin \theta = 1 \Rightarrow \sqrt{2} \sin \theta = 1 \Rightarrow \sin \theta = \frac{1}{\sqrt{2}}$.

$\theta = \frac{\pi}{4}$.

$\boxed{\text{(B)}}$

Q12

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Area of a rectangle having vertices A, B, C and D with position vectors $-i + \frac{1}{2}j + 4k$, $i + \frac{1}{2}j + 4k$, $i - \frac{1}{2}j + 4k$ and $-i - \frac{1}{2}j + 4k$, respectively is
(A) 1/2
(B) 1
(C) 2
(D) 4

$\vec{AB} = \vec{b} - \vec{a} = 2i$. $|\vec{AB}| = 2$.

$\vec{AD} = \vec{d} - \vec{a} = -j$. $|\vec{AD}| = 1$.

Area $= |\vec{AB}| \times |\vec{AD}| = 2 \times 1 = 2$.

$\boxed{\text{(C)}}$

Exercise 10.4 Practice