If the line of intersection of the planes ax | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$A$ 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:

Similar Questions

If $P=(2,-3,4)$,$Q=(-1,-4,0)$,and $R=(2,1,0)$ are three points,and $S$ is the foot of the perpendicular drawn from $R$ to the line $PQ$,then the $X$-coordinate of $S$ is:

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

$A$ line $L$ passes through the points $\hat{i}+2 \hat{j}+\hat{k}$ and $-2 \hat{i}+3 \hat{k}$. $A$ plane $P$ passes through the origin and the points $4 \hat{k}, 2 \hat{i}+\hat{j}$. The point where the line $L$ meets the plane $P$ is

The lines $\frac{x - a + d}{\alpha - \delta} = \frac{y - a}{\alpha} = \frac{z - a - d}{\alpha + \delta}$ and $\frac{x - b + c}{\beta - \gamma} = \frac{y - b}{\beta} = \frac{z - b - c}{\beta + \gamma}$ are coplanar. Find the equation of the plane in which they lie.

If a plane $P$ passes through the points $(1,0,0)$ and $(0,1,0)$ and makes an angle $\frac{\pi}{4}$ with the plane $x+y=3$,then the direction ratios of a normal to that plane $P$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module