If the line of intersection of the planes ax | vedclass

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(D) The normal vectors to the planes are $\vec{n_1} = a\hat{i} + b\hat{j} + 0\hat{k}$ and $\vec{n_2} = a\hat{i} + b\hat{j} + c\hat{k}$.The direction v

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$A$ or $B$ or both

Vedclass · Answer

(D) The normal vectors to the planes are $\vec{n_1} = a\hat{i} + b\hat{j} + 0\hat{k}$ and $\vec{n_2} = a\hat{i} + b\hat{j} + c\hat{k}$.The direction vector $\vec{v}$ of the line of intersection is given by $\vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a & b & 0 \ a & b & c \end{vmatrix} = (bc)\hat{i} - (ac)\hat{j} + 0\hat{k}$.The direction ratios of the line are proportional to $(b, -a, 0)$.The angle $	heta$ between a line with direction ratios $(l, m, n)$ and a plane with normal $\vec{N} = (A, B, C)$ is given by $\sin 	heta = \left| \frac{Al + Bm + Cn}{\sqrt{l^2+m^2+n^2} \sqrt{A^2+B^2+C^2}} ight|$.Here,the line has direction ratios $(b, -a, 0)$ and the plane $y - z + 2 = 0$ has normal $\vec{N} = (0, 1, -1)$.Given $	heta = 30^{\circ}$,so $\sin 30^{\circ} = \frac{1}{2} = \left| \frac{0(b) + 1(-a) + (-1)(0)}{\sqrt{b^2 + (-a)^2 + 0^2} \sqrt{0^2 + 1^2 + (-1)^2}} ight| = \left| \frac{-a}{\sqrt{a^2+b^2} \sqrt{2}} ight|$.Squaring both sides: $\frac{1}{4} = \frac{a^2}{2(a^2+b^2)} \Rightarrow a^2+b^2 = 2a^2 \Rightarrow b^2 = a^2 \Rightarrow b = \pm a$.If $b = a$,the direction ratios are $(a, -a, 0)$,so the direction cosines are $\pm(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)$.If $b = -a$,the direction ratios are $(-a, -a, 0)$,so the direction cosines are $\pm(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0) = \pm(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)$.Thus,the direction cosines are $\pm(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)$ or $\pm(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)$,which corresponds to options $A$ and $B$.

If the line of intersection of the planes $ax + by = 3$ and $ax + by + cz = 0$ $(a > 0)$ makes an angle $30^{\circ}$ with the plane $y - z + 2 = 0$,then the direction cosines of the line are:

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