#### Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 24.1 question 27 sub question (ii)

$\vec{a}=7(\hat{i}-\hat{j}-\hat{k})$

Hint:

To solve this question we suppose term in terms of x,y,z.

Given:

\begin{aligned} &\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} \\\\ &\vec{d} \cdot \vec{c}=21 \end{aligned}

Solution: Let $\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}$

Now, $\vec{d} \cdot \vec{c}=0$

\begin{aligned} &(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}+\hat{j}-\hat{k})=0 \\\\ &3 x+y-z=0 \end{aligned}.......(i)

Now, $\vec{d} \cdot \vec{b}=0$

\begin{aligned} &(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}-4 \hat{j}+5 \hat{k})=0 \\\\ &x-4 y+5 z=0 \end{aligned}....(ii)

Now,

\begin{aligned} &\vec{d} \cdot \vec{a}=0 \\\\ &(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(4 \hat{i}+5 \hat{j}-\hat{k})=0 \\\\ &4 x+5 y-z=0 \end{aligned}....(iii)

Solving (2) and (3)

\begin{aligned} &\frac{x}{25-4}=\frac{y}{-1-20}=\frac{z}{-16-5}=k \\\\ &\frac{x}{-21}=\frac{y}{-21}=\frac{z}{-21}=k \end{aligned}

\begin{aligned} &\frac{x}{1}=\frac{y}{-1}=\frac{z}{-1}=k \\\\ &x=1 ; y=-1 ; z=-k \end{aligned}

Putting value in (i)

\begin{aligned} &3 k-k+k=21 \\\\ &k=7 \\\\ &x=7 ; y=-7 ; z=-7 \\\\ &\vec{a}=7(\hat{i}-\hat{j}-\hat{k}) \end{aligned}