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

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

Answers (1)

Answer:

\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}

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