Let a line with direction ratios a, -4a, -7 b | vedclass

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(B) Given the line with direction ratios $(a, -4a, -7)$ is perpendicular to $(3, -1, 2b)$,the dot product is zero: $3a + 4a - 14b = 0 \implies 7a = 14

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) Given the line with direction ratios $(a, -4a, -7)$ is perpendicular to $(3, -1, 2b)$,the dot product is zero: $3a + 4a - 14b = 0 \implies 7a = 14b \implies a = 2b$ $(i)$.Also,it is perpendicular to $(b, a, -2)$,so $ab - 4a^2 + 14 = 0$ $(ii)$.Substituting $a = 2b$ into $(ii)$: $b(2b) - 4(2b)^2 + 14 = 0 \implies 2b^2 - 16b^2 + 14 = 0 \implies -14b^2 = -14 \implies b^2 = 1$.Thus,$a^2 = (2b)^2 = 4b^2 = 4$.The line equation becomes $\frac{x+1}{4+1} = \frac{y-2}{4-1} = \frac{z}{1} = k$,so $\frac{x+1}{5} = \frac{y-2}{3} = \frac{z}{1} = k$.This gives $\alpha = 5k - 1, \beta = 3k + 2, \gamma = k$.Since $(\alpha, \beta, \gamma)$ lies on $x - y + z = 0$,we have $(5k - 1) - (3k + 2) + k = 0 \implies 3k - 3 = 0 \implies k = 1$.Therefore,$\alpha + \beta + \gamma = (5k - 1) + (3k + 2) + k = 9k + 1 = 9(1) + 1 = 10$.

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