First slide

Straight lines in 3D

Question

$The coordinates of a point on the line \frac{x - 1}{2} = \frac{y + 1}{- 3} = z at a distance 4 \sqrt{14} from the point (1, - 1, 0) are$

Difficult

A

$(9, - 13, 4)$
B

$(8 \sqrt{14} + 1, - 12 \sqrt{14} - 1, 4 \sqrt{14})$
C

$(- 7, 11, - 4)$
D

$(- 8 \sqrt{14} + 1, 12 \sqrt{14} - 1, - 4 \sqrt{14})$

Solution

$Let for required point$
$\begin{array}{l} \frac{x - 1}{2} = \frac{y + 1}{- 3} = z = λ \\ \Rightarrow (1 + 2 λ, - 1 - 3 λ, λ) \\ Now distance between this point with (1, - 1, 0) is 4 \sqrt{14} \end{array}$
$\begin{array}{l} \Rightarrow (2 λ)^{2} + (- 3 λ)^{2} + λ^{2} = 16 \times 14 \\ 14 λ^{2} = 16 \times 14 \\ λ = \pm 4 \\ \Rightarrow (9, - 13, 4) and (- 7, 11, - 4) Ans. \end{array}$

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