If the equation of the line passing through the point and perpendicular to the lines and is , then a + b + c + d is equal to : [2025]

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Solution

Q.
If the equation of the line passing through the point $(0, - \frac{1}{2}, 0)$ and perpendicular to the lines $\vec{r} = λ (\hat{i} + a \hat{j} + b \hat{k})$ and $\vec{r} = (\hat{i} - \hat{j} - 6 \hat{k}) + μ (- b \hat{i} + a \hat{j} + 5 \hat{k})$ is $\frac{x - 1}{- 2} = \frac{y + 4}{d} = \frac{z - c}{- 4}$ , then a + b + c + d is equal to : [2025]

1 12

2 10

3 14

4 13

Ans.
(3)

Since, given line is perpendicular to both the line, i.e., $(\hat{i} + a \hat{j} + b \hat{k}) \times (- b \hat{i} + a \hat{j} + 5 \hat{k})$

Also, parallel vector along the required line is

$(\hat{i} + a \hat{j} + b \hat{k}) \times (- b \hat{i} + a \hat{j} + 5 \hat{k})$

$= (5 a - a b) \hat{i} - (b^{2} + 5) \hat{j} + (a + a b) \hat{k}$

$∴$ Dr's of required lines are $(5 a - a b), - (b^{2} + 5), (a + a b)$

But given Dr's of required line are –2, d, –4

$∴ \frac{5 a - a b}{- 2} = \frac{- (b^{2} + 5)}{d} = \frac{a + a b}{- 4}$ ... (i)

Since, point $(0, \frac{- 1}{2}, 0)$ lies on $\frac{x - 1}{- 2} = \frac{y + 4}{d} = \frac{z - c}{- 4}$ ,

$\Rightarrow \frac{0 - 1}{- 2} = \frac{\frac{- 1}{2} + 4}{d} = \frac{0 - c}{- 4} \Rightarrow d = 7, c = 2$

From (i), $\frac{5 a - a b}{- 2} = \frac{- b^{2} - 5}{7} = \frac{a + a b}{- 4}$

$\begin{matrix} \frac{5 a - a b}{- 2} = \frac{a + a b}{- 4} \\ \Rightarrow - 20 a + 4 a b = - 2 a - 2 a b \\ \Rightarrow 18 a = 6 a b \\ \Rightarrow b = 3 \end{matrix} |$ $\begin{matrix} \frac{- b^{2} - 5}{7} = \frac{a + a b}{- 4} \\ \Rightarrow 4 b^{2} + 20 = 7 a + 7 a b \\ \Rightarrow 36 + 20 = 7 a + 21 a \\ \Rightarrow 56 = 28 a \Rightarrow a = 2 \end{matrix}$

$∴$ a + b + c + d = 2 + 3 + 2 + 7 = 14

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