Let <img alt="\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}" src="https://entrancecorner.codecogs.com/gif.latex?%5Cvec%7Ba%7D%3D%204%5Chat%7Bi%7D+5%5Chat%7Bj%7D-%5Chat%7Bk%7D%2C

Let $\vec{a}= 4\hat{i}+5\hat{j}-\hat{k},\vec{b}= \hat{i}-4\hat{j}+5\hat{k}\; and\; \vec{c}= 3\hat{i}+\hat{j}-\hat{k}$ .Find a vector $\vec{d}$ which is perpendicular to both $\vec{c}$ and $\vec{b}$ and $\vec{d}\cdot \vec{a}= 21\cdot$

Answers (1)

given $\vec{a}= 4\hat{i}+5\hat{j}-\hat{k},\vec{b}= \hat{i}-4\hat{j}+5\hat{k}, \vec{c}= 3\hat{i}+\hat{j}-\hat{k}$
Let $\vec{d}= x\hat{i}+y\hat{j}+z\hat{k}$
$\because \vec{d}$ is perpendicular to $\vec{c}$ and $\vec{b}$
$\therefore \vec{d}\cdot \vec{b}= 0$
$x-4y+5z= 0---(i)$
and $\vec{d}\cdot \vec{c}= 0$
$3x+y-z= 0---(ii)$
Also $\vec{d}\cdot \vec{a}= 21 \: \: \: \left [ given \right ]$
$\Rightarrow 5y+4x-z= 21---(iii)$
On substracting equation (ii) from equation (iii) we get
$x+4y= 21---(iv)$
On multiplying equation (iii) by 5 and adding equation (i) we get.
$21x+21y= 105$
$x+y= 5---(v)$
On substracting equation (v) from equation (iv), we get
$3y= 16\Rightarrow y= \frac{16}{3}$
From equation (v)
$x= 5-\frac{16}{3}= \frac{-1}{3}$
From equation (ii)
$z= 3x+y$
$= 3\left ( \frac{-1}{3} \right )+\frac{16}{3}= \frac{13}{3}$
$\therefore \vec{d}= \frac{-1}{3}\hat{i}+\frac{16}{3}\hat{j}+\frac{13}{3}\hat{k}$