First slide

Straight lines in 3D

Question

The point on the line $\frac{x + 2}{2} = \frac{y + 6}{3} = \frac{z - 34}{- 10}$ which is nearest to the line $\frac{x + 6}{4} = \frac{y - 7}{- 3} = \frac{z - 7}{- 2}$ is $(a, b, c)$ where $a + b + c =$

Moderate

A

9
B

10
C

11
D

12

Solution

The given two lines are not parallel lines, so both possess common perpendicular line. The ends of this common perpendicular are the nearest points to each other
The general point on the line $\frac{x + 2}{2} = \frac{y + 6}{3} = \frac{z - 34}{- 10}$ is $P (- 2 + 2 s, - 6 + 3 s, 34 - 10 s)$
And the general point on the line $\frac{x + 6}{4} = \frac{y - 7}{- 3} = \frac{z - 7}{- 2}$ is $Q (- 6 + 4 t, 7 - 3 t, 7 - 2 t)$
The direction ration of line joining $P, Q$ is $〈- 4 + 4 t - 2 s, 13 - 3 t - 3 s, - 27 - 2 t + 10 s〉$
The line $P Q$ is perpendicular to both lines
Hence,
$\begin{matrix} 2 (- 4 + 4 t - 2 s) + 3 (13 - 3 t - 3 s) - 10 (- 27 - 2 t + 10 s) = 0 \\ 4 (- 4 + 4 t - 2 s) - 3 (13 - 3 t - 3 s) - 2 (- 27 - 2 t + 10 s) = 0 \end{matrix}$
It gives, $s = 3, t = 2$
The point $P (4, 3, 4)$ is required point.
Therefore, $a + b + c = 4 + 3 + 4 = 11$

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