The equation of the line passing through the | vedclass

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Question

(A) Let the direction ratios of the required line be $(l, m, n)$.Since the line is perpendicular to the lines with direction ratios $(3, -16, 7)$ and

Vedclass · Accepted Answer

$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$

Vedclass · Answer

(A) Let the direction ratios of the required line be $(l, m, n)$.Since the line is perpendicular to the lines with direction ratios $(3, -16, 7)$ and $(3, 8, -5)$,we have:$3l - 16m + 7n = 0$ $(i)$$3l + 8m - 5n = 0$ (ii)Subtracting (ii) from $(i)$,we get: $-24m + 12n = 0 \implies 24m = 12n \implies n = 2m$.Substituting $n = 2m$ in $(i)$: $3l - 16m + 7(2m) = 0 \implies 3l - 16m + 14m = 0 \implies 3l = 2m \implies l = \frac{2}{3}m$.Taking $m = 3$,we get $l = 2$ and $n = 6$.Thus,the direction ratios are $(2, 3, 6)$.The equation of the line passing through $(1, 2, -4)$ with direction ratios $(2, 3, 6)$ is $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$.

The equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5}$ is:

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