If the direction ratios a, b, c of a line L s | vedclass

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Question

(D) Given the equations: $ab + bc + ca = 0$  --- $(1)$ $6ab + 9bc + 8ca = 0$ --- $(2)$ Multiply equation $(1)$ by $6$: $6ab + 6bc + 6ca = 0$ --- $(3)$

Vedclass · Accepted Answer

$\frac{-3}{7}, \frac{2}{7}, \frac{-6}{7}$

Vedclass · Answer

(D) Given the equations: $ab + bc + ca = 0$  --- $(1)$ $6ab + 9bc + 8ca = 0$ --- $(2)$ Multiply equation $(1)$ by $6$: $6ab + 6bc + 6ca = 0$ --- $(3)$ Subtracting $(3)$ from $(2)$: $(6ab + 9bc + 8ca) - (6ab + 6bc + 6ca) = 0$ $3bc + 2ca = 0 \Rightarrow c(3b + 2a) = 0$. Since $c 
eq 0$ (otherwise $a=b=0$, which is not possible for direction ratios), we get $2a = -3b$, or $a/(-3) = b/2$. Substitute $a = -3k$ and $b = 2k$ into equation $(1)$: $(-3k)(2k) + (2k)c + c(-3k) = 0$ $-6k^2 - kc = 0 \Rightarrow c = -6k$. Thus, the direction ratios are proportional to $(-3k, 2k, -6k)$, i.e.,$(-3, 2, -6)$. The magnitude is $\sqrt{(-3)^2 + 2^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$. Therefore, the direction cosines are $\frac{-3}{7}, \frac{2}{7}, \frac{-6}{7}$.

If the direction ratios $a, b, c$ of a line $L$ satisfy the relations $ab + bc + ca = 0$ and $6ab + 9bc + 8ca = 0$, then the direction cosines of the line $L$ are

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