If the lines- 2-x 3 - 2y-3 3k - z-4 2 and 2-2x 5k - y-4 2 - 5-z 4 are perpendicular then find the value of k-

The direction vectors of the lines represented in standard form x 2 3=y 1.51.5k=z 42 and x 1 2.5k=y+42=z 5 4 are n⃗_⃗1⃗= 3î+1.5kĵ+2k̂ n⃗_⃗2 ...

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Byju's Answer

Standard XII

Physics

Relative Velocity

If the lines:...

Question

If the lines:
$\frac{2 - x}{3} = \frac{2 y - 3}{3 k} = \frac{z - 4}{2}$ and $\frac{2 - 2 x}{5 k} = \frac{y + 4}{2} = \frac{5 - z}{4}$ are perpendicular then find the value of $k$ .

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Solution

The direction vectors of the lines represented in standard form $\frac{x - 2}{- 3} = \frac{y - 1.5}{1.5 k} = \frac{z - 4}{2}$ and $\frac{x - 1}{- 2.5 k} = \frac{y + 4}{2} = \frac{z - 5}{- 4}$ are
$_{1} = - 3^i + 1.5 k^j + 2^k$
$_{2} = - 2.5 k^i + 2^j - 4^k$
Now, as the lines are perpendicular, their dot product is 0
$∴_{1} \cdot_{2} = 0$
$∴ (- 3^i + 1.5 k^j + 2^k) \cdot (- 2.5 k^i + 2^j - 4^k) = 0$
$∴ 7.5 k + 3 k - 8 = 0$
$∴ 10.5 k = 8$
$∴ k = \frac{16}{21}$
This is the required value of $k$ .

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If the lines:2−x3=2y−33k=z−42 and 2−2x5k=y+42=5−z4 are perpendicular then find the value of k.

If the lines:
$\frac{2 - x}{3} = \frac{2 y - 3}{3 k} = \frac{z - 4}{2}$ and $\frac{2 - 2 x}{5 k} = \frac{y + 4}{2} = \frac{5 - z}{4}$ are perpendicular then find the value of $k$ .