7 Vector equation of the line 5x+4=3y−5=−3z−8 is (a)r=4^−5^−8k^+μ(5^+3^−3k^) (A) r=−4^+5^+8k^+μ(5^+3^−3k^) (B) r=5^+3^−3k^+μ(4^−5^−8k^) (C) r=5^+3^−3k^+μ(−4^+5^−8k)

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7 Vector equation of the line {x+4}{5}={y-5}{3}={z-8}{-3} is (

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7 Vector equation of the line $\frac{x + 4}{5} = \frac{y - 5}{3} = \frac{z - 8}{- 3}$ is $(a) r = 4 \overset{}{^} - 5 \overset{}{^} - 8 \hat{k} + μ (5 \overset{}{^} + 3 \overset{}{^} - 3 \hat{k})$

$r = - 4 \overset{}{^} + 5 \overset{}{^} + 8 \hat{k} + μ (5 \overset{}{^} + 3 \overset{}{^} - 3 \hat{k})$
$r = 5 \overset{}{^} + 3 \overset{}{^} - 3 \hat{k} + μ (4 \overset{}{^} - 5 \overset{}{^} - 8 \hat{k})$
$r = 5 \overset{}{^} + 3 \overset{}{^} - 3 \hat{k} + μ (- 4 \overset{}{^} + 5 \overset{}{^} - 8 k)$

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Updated on: Feb 25, 2023

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7 Vector equation of the line

\frac{x + 4}{5} = \frac{y - 5}{3} = \frac{z - 8}{- 3}

(a) r = 4 \overset{}{^} - 5 \overset{}{^} - 8 \hat{k} + μ (5 \overset{}{^} + 3 \overset{}{^} - 3 \hat{k})

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Question Text	7 Vector equation of the line $\frac{x + 4}{5} = \frac{y - 5}{3} = \frac{z - 8}{- 3}$ is $(a) r = 4 \overset{}{^} - 5 \overset{}{^} - 8 \hat{k} + μ (5 \overset{}{^} + 3 \overset{}{^} - 3 \hat{k})$
Updated On	Feb 25, 2023
Topic	Coordinate Geometry
Subject	Mathematics
Class	Class 12