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  • Provide Solution for RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Very Short Answer Question 8

Provide Solution for RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Very Short Answer Question 8

Answers (1)

Answer:

The given line is perpendicular to z-axis

Hint:

Use properties of vector

Given:

\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-1}{0}

Solution:

We have

\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-1}{0}

The given line is parallel to the vector  \vec{b}=3 \hat{i}+4 \hat{j}+0 \hat{k}

Let,  x \hat{i}+y \hat{j}+z \hat{k}  be perpendicular to the given line

Now,  3 x+4 y+0 z=0

Direction ratios are  \left(a_{1}, b_{1}, c_{1}\right)=(3,4,0)

Check perpendicular condition with x axis

Direction ratio of x axis is  \left(a_{2}, b_{2}, c_{2}\right)=(1,0,0)

\begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2} \\ & \end{aligned}

=3(1)+4(0)+0(0) \\

=3

\neq 0   Hence x axis is not perpendicular

 Check perpendicular condition with y axis

Direction ratio of y axis is  \left(a_{2}, b_{2}, c_{2}\right)=(0,1,0)

\begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2} \\ & \end{aligned}

=3(0)+4(1)+0(0) \\

=4 \neq 0

Hence y axis is not perpendicular

Check perpendicular condition with z axis

Direction ratio of z axis is  \left(a_{2}, b_{2}, c_{2}\right)=(0,0,1)

\begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2} \\ & \end{aligned}

=3(0)+4(0)+0(1) \\

=0

Hence, the given line is perpendicular to the z-axis

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