The line $\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z - k}{3}$ cuts the $Y Z$ and $Z X$ planes at $A$ and $B$ respectively. If $∠ A O B = \frac{π}{2}$ then $k$ is:

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Solution

Any point on the given line be $(r - 1, 2 r + 2, 3 r + k)$ .
This point will lie on the $Y Z$ plane if $r - 1 = 0 \Rightarrow r = 1$ .
So, co-ordinate of $A$ is $(0, 4, 3 + k)$ .
Also, that point will lie on $Z X$ plane if $2 r + 2 = 0 \Rightarrow r = - 1$ . So, co-ordinate of $B$ is $(- 2, 0, k - 3)$ .
In vectorial notion $\to O A = 4^j + (k + 3)^k$ and $\to O B = - 2^i + (k - 3)^k$ , where $O$ is the origin.
Now, angle between $\to O A$ and $\to O B$ $= {cos}^{- 1} (\frac{\to O A . \to O B}{| \to O A | | \to O B |})$ .
Now, according to the problem,
$\to O A . \to O B = 0$
$\Rightarrow k^{2} - 9 = 0$
$\Rightarrow k = \pm 3$

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The line x+11=y−22=z−k3 cuts the YZ and ZX planes at A and B respectively. If ∠AOB=π2 then k is:

The line $\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z - k}{3}$ cuts the $Y Z$ and $Z X$ planes at $A$ and $B$ respectively. If $∠ A O B = \frac{π}{2}$ then $k$ is: