In the previous article, we discussed Very Short Answer Type Questions (Q11 to Q17) from the Chapter Product of Two Vectors of Chhaya Mathematics (Class XII) / S N Dey Mathematics Class 12. In this article, we will solve Short Answer Type Questions from the book.
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$ (3\vec{a}-5\vec{b})\cdot (2\vec{a}+7\vec{b}) $
$ = 6\vec{a}\cdot \vec{a}+21 \vec{a}\cdot \vec{b}-10\vec{b}\cdot \vec{a}-35 \vec{b}\cdot \vec{b} $
$ = 6|\vec{a}|^2+11\vec{a}\cdot \vec{b}-35|\vec{b}|^2 $
$ = 6 \times 2^2+11 \times 1-35 \times 1^2 $
$ = 24+11-35 = 0 ~~\text{(Ans.)} $
The projection of $~~\vec{b}~~$ on $~~\vec{a}~~$ is given by:
$ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} = \frac 32 ~~\text{(Ans.)} $
According to the problem,
$ \frac{(\hat{i}+\hat{j}) \cdot (\lambda \hat{i}-\hat{j})}{|\hat{i}+\hat{j}|} = 0 $
$ \implies \frac{\lambda (\hat{i} \cdot \hat{i})-\hat{i} \cdot \hat{j}+\lambda (\hat{j} \cdot \hat{i})-\hat{j} \cdot \hat{j}}{\sqrt{1^2+1^2}} = 0 $
$ \implies \lambda - 1 = 0 \quad [\because \hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{i}=0] $
$ \therefore \lambda = 1 ~~\text{(Ans.)} $
Since $ (\vec{a}+\lambda \vec{b}) \perp (\vec{a}-\lambda \vec{b}) $:
$ (\vec{a}+\lambda \vec{b}) \cdot (\vec{a}-\lambda \vec{b}) = 0 $
$ \implies \vec{a} \cdot \vec{a}-\lambda \vec{a} \cdot \vec{b}+\lambda \vec{b} \cdot \vec{a}-\lambda^2 \vec{b} \cdot \vec{b} = 0 $
$ \implies |\vec{a}|^2-\lambda^2|\vec{b}|^2 = 0 \quad [\because \vec{b} \cdot \vec{a}=\vec{a} \cdot \vec{b}] $
$ \implies 4^2-\lambda^2 \times 3^2 = 0 \implies 16 = 9\lambda^2 $
$ \therefore \lambda = \pm \sqrt{\frac{16}{9}} = \pm \frac 43 ~~\text{(Ans.)} $
Since $\vec{a}+\vec{b}+\vec{c}=\vec{0}$:
$ (\vec{a}+\vec{b}+\vec{c})\cdot (\vec{a}+\vec{b}+\vec{c}) = 0 $
$ \implies |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = 0 $
$ \implies 3^2+4^2+5^2 + 2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = 0 $
$ \implies 50 + 2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = 0 $
$ \therefore \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} = -25 ~~\text{(Proved)} $
Let $~~\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}$
$ \vec{a} \cdot \vec{r}=1 \implies x-y+2z=1 \quad \text{--- (1)} $
$ \vec{b} \cdot \vec{r}=2 \implies x+y+z=2 \quad \text{--- (2)} $
$ \vec{c} \cdot \vec{r}=5 \implies 2x-y+z=5 \quad \text{--- (3)} $
Adding (1) and (2): $ 2x+3z=3 \quad \text{--- (4)} $
Adding (2) and (3): $ 3x+2z=7 \quad \text{--- (5)} $
Solving (4) and (5), we get $ x = 3, z = -1 $.
Substituting into (2): $ 3 + y - 1 = 2 \implies y = 0 $.
$ \therefore \vec{r} = 3\hat{i} - \hat{k} ~~\text{(Ans.)} $
(i) $ |\vec{\alpha}+\vec{\beta}|^2 = |\vec{\alpha}|^2+|\vec{\beta}|^2 $
(ii) $ |\vec{\alpha}+\vec{\beta}|^2 = |\vec{\alpha}-\vec{\beta}|^2 $
Since $\vec{\alpha} \perp \vec{\beta}$, $ \vec{\alpha} \cdot \vec{\beta} = 0 = \vec{\beta} \cdot \vec{\alpha} \quad \text{--- (1)} $
(i) $ |\vec{\alpha}+\vec{\beta}|^2 = (\vec{\alpha}+\vec{\beta}) \cdot (\vec{\alpha}+\vec{\beta}) $
$ = \vec{\alpha} \cdot \vec{\alpha} + \vec{\alpha} \cdot \vec{\beta} + \vec{\beta} \cdot \vec{\alpha} + \vec{\beta} \cdot \vec{\beta} = |\vec{\alpha}|^2 + |\vec{\beta}|^2 ~~\text{(Proved)} $
(ii) $ |\vec{\alpha}+\vec{\beta}|^2 - |\vec{\alpha}-\vec{\beta}|^2 $
$ = (|\vec{\alpha}|^2+|\vec{\beta}|^2+2\vec{\alpha}\cdot \vec{\beta}) - (|\vec{\alpha}|^2+|\vec{\beta}|^2-2\vec{\alpha} \cdot \vec{\beta}) $
$ = 4\vec{\alpha} \cdot \vec{\beta} = 0 \quad [\text{By (1)}] $
$ \therefore |\vec{\alpha}+\vec{\beta}|^2 = |\vec{\alpha}-\vec{\beta}|^2 ~~\text{(Proved)} $
The position vectors are $A = p\hat{i}+\hat{j}-\hat{k}$, $B = 2p\hat{i}+2\hat{k}$, $C = (2+2p)\hat{i}+p\hat{j}+p\hat{k}$.
$ \vec{AB} = (2p-p)\hat{i} - \hat{j} + (2+1)\hat{k} = p\hat{i}-\hat{j}+3\hat{k} $
$ \vec{BC} = (2+2p-2p)\hat{i} + p\hat{j} + (p-2)\hat{k} = 2\hat{i}+p\hat{j}+(p-2)\hat{k} $
Since $ \vec{AB} \perp \vec{BC} \implies \vec{AB} \cdot \vec{BC} = 0 $:
$ (p\hat{i}-\hat{j}+3\hat{k}) \cdot [2\hat{i}+p\hat{j}+(p-2)\hat{k}] = 0 $
$ \implies 2p - p + 3(p-2) = 0 \implies 4p - 6 = 0 \implies p = \frac 32 ~~\text{(Ans.)} $
$ 2\vec{\alpha}+\vec{\beta} = 2(\hat{i}+2\hat{j}-3\hat{k}) + (3\hat{i}-\hat{j}+2\hat{k}) = 5\hat{i}+3\hat{j}-4\hat{k} $
$ |2\vec{\alpha}+\vec{\beta}| = \sqrt{5^2+3^2+(-4)^2} = \sqrt{50} $
$ \vec{\alpha}+2\vec{\beta} = (\hat{i}+2\hat{j}-3\hat{k}) + 2(3\hat{i}-\hat{j}+2\hat{k}) = 7\hat{i}+\hat{k} $
$ |\vec{\alpha}+2\vec{\beta}| = \sqrt{7^2+1^2} = \sqrt{50} $
$ (2\vec{\alpha}+\vec{\beta}) \cdot (\vec{\alpha}+2\vec{\beta}) = (5)(7) + (3)(0) + (-4)(1) = 31 $
$ \cos\theta = \frac{31}{\sqrt{50}\sqrt{50}} = \frac{31}{50} ~~\text{(Ans.)} $
$ \vec{a} \cdot \left(\vec{b}-\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}~\vec{a}\right) = \vec{a} \cdot \vec{b} - \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}(\vec{a} \cdot \vec{a}) $
$ = \vec{a} \cdot \vec{b} - \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} |\vec{a}|^2 = \vec{a} \cdot \vec{b} - \vec{a}\cdot \vec{b} = 0 $
Since the dot product is zero, the vectors are perpendicular. $~~\text{(Proved)}$
If $\theta$ is the angle between $\vec{a}$ and $\vec{b}$:
$ \vec{a}\cdot \vec{b} = |\vec{a}||\vec{b}| \cos\theta \implies \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \quad \text{--- (1)} $
We know that $ |\cos\theta| \leq 1 \quad \text{--- (2)} $
Substituting (1) into (2): $ \left|\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right| \leq 1 \implies |\vec{a} \cdot \vec{b}| \leq |\vec{a}||\vec{b}| ~~\text{(Proved)} $
Let $ \vec{r} = x\vec{a}+y\vec{b}+z\vec{c} $ where $x, y, z$ are scalars.
$ \vec{r} \cdot \vec{a} = 0 \implies x|\vec{a}|^2 + y(\vec{a} \cdot \vec{b}) + z(\vec{a} \cdot \vec{c}) = 0 \quad \text{--- (1)} $
$ \vec{r} \cdot \vec{b} = 0 \implies x(\vec{a} \cdot \vec{b}) + y|\vec{b}|^2 + z(\vec{b} \cdot \vec{c}) = 0 \quad \text{--- (2)} $
$ \vec{r} \cdot \vec{c} = 0 \implies x(\vec{a} \cdot \vec{c}) + y(\vec{b} \cdot \vec{c}) + z|\vec{c}|^2 = 0 \quad \text{--- (3)} $
Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, the determinant of coefficients $[\vec{a}~\vec{b}~\vec{c}] \neq 0$. Thus, the only solution is $ x = y = z = 0 \implies \vec{r} = \vec{0} ~~\text{(Proved)} $.
$ |\vec{a}+\vec{b}|^2 = |\vec{a}-\vec{b}|^2 $
$ \implies (\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = (\vec{a}-\vec{b}) \cdot (\vec{a}-\vec{b}) $
$ \implies |\vec{a}|^2+|\vec{b}|^2+2\vec{a} \cdot \vec{b} = |\vec{a}|^2+|\vec{b}|^2-2\vec{a} \cdot \vec{b} $
$ \implies 4\vec{a} \cdot \vec{b} = 0 \implies \vec{a} \cdot \vec{b} = 0 \implies \vec{a} \perp \vec{b} ~~\text{(Proved)} $
$ (\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = |\vec{a}|^2+|\vec{b}|^2 $
$ \implies |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2 = |\vec{a}|^2+|\vec{b}|^2 $
$ \implies 2\vec{a} \cdot \vec{b} = 0 \implies \vec{a} \cdot \vec{b} = 0 \implies \vec{a} \perp \vec{b} ~~\text{(Proved)} $
(i) $ \vec{a} \cdot (\vec{b}+\vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} $
(ii) $ (\vec{b}+\vec{c}) \times \vec{a} = \vec{b} \times \vec{a} + \vec{c} \times \vec{a} $
(i) $ \vec{b}+\vec{c} = (b_1+c_1)\hat{i} + (b_2+c_2)\hat{j} + (b_3+c_3)\hat{k} $
$ \vec{a} \cdot (\vec{b}+\vec{c}) = a_1(b_1+c_1) + a_2(b_2+c_2) + a_3(b_3+c_3) $
$ = (a_1b_1+a_2b_2+a_3b_3) + (a_1c_1+a_2c_2+a_3c_3) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} ~~\text{(Proved)} $
(ii) $ (\vec{b}+\vec{c}) \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ (b_1+c_1) & (b_2+c_2) & (b_3+c_3) \\ a_1 & a_2 & a_3 \end{vmatrix} $
$ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \end{vmatrix} + \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ c_1 & c_2 & c_3 \\ a_1 & a_2 & a_3 \end{vmatrix} = \vec{b} \times \vec{a} + \vec{c} \times \vec{a} ~~\text{(Proved)} $
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