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(B) The required line is perpendicular to two lines with direction vectors $\vec{v_1} = \hat{i} + a\hat{j} + b\hat{k}$ and $\vec{v_2} = -b\hat{i} + a\

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) The required line is perpendicular to two lines with direction vectors $\vec{v_1} = \hat{i} + a\hat{j} + b\hat{k}$ and $\vec{v_2} = -b\hat{i} + a\hat{j} + 5\hat{k}$.Thus,the direction vector of the required line is $\vec{v} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & a & b \ -b & a & 5 \end{vmatrix} = \hat{i}(5a - ab) - \hat{j}(5 + b^2) + \hat{k}(a + ab)$.The direction ratios of the line $\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4}$ are $(-2, d, -4)$.Since the lines are parallel,their direction ratios are proportional: $\frac{5a - ab}{-2} = \frac{-(b^2 + 5)}{d} = \frac{a + ab}{-4} = k$.Since the point $(0, -\frac{1}{2}, 0)$ lies on the line $\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4}$,we have $\frac{0-1}{-2} = \frac{-1/2 + 4}{d} = \frac{0-c}{-4} \Rightarrow \frac{1}{2} = \frac{7/2}{d} = \frac{-c}{-4}$.From $\frac{1}{2} = \frac{7}{2d}$,we get $d = 7$. From $\frac{1}{2} = \frac{c}{4}$,we get $c = 2$.Now,$\frac{5a - ab}{-2} = \frac{a + ab}{-4} \Rightarrow 2(5a - ab) = a + ab \Rightarrow 10a - 2ab = a + ab \Rightarrow 9a = 3ab \Rightarrow b = 3$.Also,$\frac{-(b^2 + 5)}{d} = \frac{a + ab}{-4} \Rightarrow \frac{-(9 + 5)}{7} = \frac{a + 3a}{-4} \Rightarrow -2 = \frac{4a}{-4} \Rightarrow -2 = -a \Rightarrow a = 2$.Thus,$a+b+c+d = 2 + 3 + 2 + 7 = 14$.

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