The distance of line $3y - 2z - 1 = 0 = 3x - z + 4$ from the point (2, $-$1, 6) is :
Solution
$3y - 2z - 1 = 0 = 3x - z + 4$<br><br>$3y - 2z - 1 = 0$<br><br>D.R's $\Rightarrow$ (0, 3, $-$2)<br><br>$3x - z + 4$ = 0<br><br>D.R's $\Rightarrow$ (3, $-$1, 0)<br><br>Let DR's of given line are a, b, c<br><br>Now, 3b $-$ 2c = 0 & 3a $-$ c = 0<br><br>$\therefore$ 6a = 3b = 2c<br><br>a : b : c = 3 : 6 : 9<br><br>Any point on line<br><br>3K $-$ 1, 6K + 1, 9K + 1<br><br>Now, 3(3K $-$ 1) + 6(6K + 1)1 + 9(9K + 1) = 0<br><br>$\Rightarrow$ K = ${1 \over 3}$<br><br>Point on line $\Rightarrow$ (0, 3, 4)<br><br>Given point (2, $-$1, 6)<br><br>$\Rightarrow$ Distance = $\sqrt {4 + 16 + 4} = 2\sqrt 6$<br><br>Option (c)
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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