In the previous article, we discussed Very Short Answer Type Questions (Q1 to Q10) from the Chapter Product of Two Vectors of Chhaya Mathematics (Class XII) / S N Dey Mathematics Class 12. In this article, we will solve additional problems related to Vector Product.
.png)
$ \vec{b}+\vec{c} = (\hat{i}+2\hat{j}-2\hat{k})+(2\hat{i}-\hat{j}+4\hat{k}) = 3\hat{i}+\hat{j}+2\hat{k} $
$ (\vec{b}+\vec{c}) \cdot \vec{a} = (3\hat{i}+\hat{j}+2\hat{k}) \cdot (2\hat{i}-2\hat{j}+\hat{k}) = 6-2+2 = 6 $
$ |\vec{a}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{9}=3 $
The projection of $(\vec{b}+\vec{c})$ on vector $\vec{a}$ is given by:
$ =\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=\frac 63=2~~\text{unit} $
$ \vec{b}+\vec{c} = (\hat{i}+3\hat{j}+\hat{k})+(\hat{i}+\hat{k}) = 2\hat{i}+3\hat{j}+2\hat{k} $
$ (\vec{b}+\vec{c}) \cdot \vec{a} = (2\hat{i}+3\hat{j}+2\hat{k}) \cdot (\hat{i}+2\hat{j}+\hat{k}) = 2+6+2 = 10 $
$ |\vec{a}| = \sqrt{1^2+2^2+1^2} = \sqrt{6} $
The projection of $(\vec{b}+\vec{c})$ on vector $\vec{a}$ is given by:
$ =\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|} = \frac{10}{\sqrt{6}} = \frac{10\sqrt{6}}{6} = \frac{5\sqrt{6}}{3}~~\text{unit} $
If $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$, then:
$ \cos\theta = \frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{\sqrt{6}}{\sqrt{3} \cdot 2} = \frac{\sqrt{3} \sqrt{2}}{2\sqrt{3}} = \frac{1}{\sqrt{2}} = \cos(\pi/4) $
$ \therefore~\theta=\frac{\pi}{4} ~~\text{(Ans.)} $
Let $\vec{a}=\sqrt{2} \hat{i}+\hat{j}+\hat{k}$. The unit vector along $y$-axis is $\vec{b}=\hat{j}$.
$ \vec{a} \cdot \vec{b} = (\sqrt{2} \hat{i}+\hat{j}+\hat{k}) \cdot (\hat{j}) = 1 $
$ |\vec{a}| = \sqrt{(\sqrt{2})^2+1^2+1^2} = \sqrt{4} = 2, \quad |\vec{b}| = 1 $
If $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then:
$ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{1}{2 \times 1} = \frac 12 = \cos(\pi/3) $
$ \therefore~ \theta=\frac{\pi}{3}~~\text{(Ans.)} $
$ |\vec{a}-\vec{b}|^2 = (\vec{a}-\vec{b}) \cdot (\vec{a}-\vec{b}) = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 $
$ = 2^2 - 2(4) + 3^2 = 4 - 8 + 9 = 5 $
$ \therefore~|\vec{a}-\vec{b}|=\sqrt{5}~~\text{(Ans.)} $
Refer to text: Chhaya Mathematics (S N Dey)
If $\theta$ is the angle between vectors $\vec{a}$ and $\vec{b}$, then:
$ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin\theta \implies 6 = 3 \times 4 \times \sin\theta $
$ \sin\theta = \frac{6}{12} = \frac 12 = \sin(\pi/6) $
$ \therefore~ \theta=\frac{\pi}{6}~~\text{(Ans.)} $
$ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin\theta \implies 1 = 1 \times 1 \times \sin\theta $
$ \sin\theta = 1 = \sin(\pi/2) \implies \theta = \frac{\pi}{2}~~\text{(Ans.)} $
$ |\vec{a}||\vec{b}| \cos\theta = |\vec{a}||\vec{b}|\sin\theta \implies \cos\theta = \sin\theta $
$ \frac{\sin\theta}{\cos\theta} = 1 \implies \tan\theta = 1 = \tan(\pi/4) \implies \theta = \frac{\pi}{4}~~\text{(Ans.)} $
$ \sin\theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|} = \frac{1}{\sqrt{3} \times \frac 23} = \frac{\sqrt{3}}{2} $
$ \sin\theta = \sin(\pi/3) \implies \theta = \frac{\pi}{3}~~\text{(Ans.)} $
Two vectors are parallel if $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}$. Therefore:
$ \frac{p}{-3} = \frac 84 = \frac 6q \quad \text{--- (1)} $
From (1): $\frac{p}{-3} = 2 \implies p = -6$
From (1): $\frac 6q = 2 \implies q = 3$
$ \begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 2& 6 & 27 \\ 1& 3 & p \end{vmatrix} = \vec{0} $
$ (6p-81)\hat{i} - (2p-27)\hat{j} + (6-6)\hat{k} = \vec{0} $
$ 2p-27 = 0 \implies p = \frac{27}{2}~~\text{(Ans.)} $
$ \begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 2& 6 & 14 \\ 1& -\lambda & 7 \end{vmatrix} = \vec{0} $
$ (42+14\lambda)\hat{i} - (2\lambda+6)\hat{k} = \vec{0} $
$ 2\lambda+6 = 0 \implies \lambda = -3~~\text{(Ans.)} $
- (i) adjacent sides are $ \vec{a}=3\hat{i}+\hat{j}+4\hat{k} $ and $ \vec{b}=\hat{i}-\hat{j}+\hat{k} $.
- (ii) vertices are $(0,-3,-1), (2,1,-1), (3,-3,2)$ and $(1,-7,2)$ taken in order.
- (iii) diagonals are the vectors $ 3\hat{i}+\hat{j}-2\hat{k} $ and $ \hat{i}-3\hat{j}+4\hat{k} $.
(i) $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 3& 1 & 4 \\ 1& -1 & 1 \end{vmatrix} = 5\hat{i}+\hat{j}-4\hat{k}$
Area = $|\vec{a} \times \vec{b}| = \sqrt{5^2+1^2+(-4)^2} = \sqrt{42}~~\text{sq. units}$
(ii) Let $A = -3\hat{j}-\hat{k}, B = 2\hat{i}+\hat{j}-\hat{k}, C = 3\hat{i}-3\hat{j}+2\hat{k}, D = \hat{i}-7\hat{j}+2\hat{k}$.
$\vec{AB} = 2\hat{i}+4\hat{j}, \quad \vec{AD} = \hat{i}-4\hat{j}+3\hat{k}$
$\vec{AB} \times \vec{AD} = 12\hat{i}-6\hat{j}-12\hat{k}$
Area = $|\vec{AB} \times \vec{AD}| = \sqrt{144+36+144} = 18~~\text{sq. units}$
(iii) Let $\vec{a}=3\hat{i}+\hat{j}-2\hat{k}, \vec{b}=\hat{i}-3\hat{j}+4\hat{k}$.
$\vec{a} \times \vec{b} = -2\hat{i}-14\hat{j}-10\hat{k} \implies |\vec{a} \times \vec{b}| = \sqrt{300} = 10\sqrt{3}$
Area = $\frac 12|\vec{a} \times \vec{b}| = 5\sqrt{3}~~\text{sq. units}$
- (i) drawn on vectors $ \vec{a}=6\hat{i}+2\hat{j}-3\hat{k} $ and $ \vec{b}=4\hat{i}-\hat{j}-2\hat{k} $.
- (ii) whose vertices have position vectors $ \hat{i}+\hat{j}+2\hat{k}, 2\hat{i}+2\hat{j}+3\hat{k} $ and $ 3\hat{i}-\hat{j}-\hat{k} $.
- (iii) whose vertices are $(1,2,3), (2,3,1),$ and $(1,1,1)$.
(i) $\vec{a} \times \vec{b} = -7\hat{i}-14\hat{k}$
Area = $\frac 12|\vec{a} \times \vec{b}| = \frac 72\sqrt{(-1)^2+(-2)^2} = \frac{7\sqrt{5}}{2}~~\text{sq. units}$
(ii) $\vec{AB} = \hat{i}+\hat{j}+\hat{k}, \quad \vec{AC} = 2\hat{i}-2\hat{j}-3\hat{k}$
$\vec{AB} \times \vec{AC} = -\hat{i}+5\hat{j}-4\hat{k}$
Area = $\frac 12|\vec{AB} \times \vec{AC}| = \frac 12\sqrt{42} = \sqrt{\frac{21}{2}}~~\text{sq. units}$
(iii) $\vec{AB} = \hat{i}+\hat{j}-2\hat{k}, \quad \vec{AC} = -\hat{j}-2\hat{k}$
$\vec{AB} \times \vec{AC} = -4\hat{i}+2\hat{j}-\hat{k}$
Area = $\frac 12|\vec{AB} \times \vec{AC}| = \frac 12\sqrt{21}~~\text{sq. units}$
Thanks for publishing this
ReplyDeletePlease do not enter any spam link in the comment box