The angle between the line r = (i + 2j - k) + | vedclass

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

Question

(A) The line is given by $r = (i + 2j - k) + \lambda (i - j + k)$. The direction vector of the line is $b = i - j + k$.The plane is given by $r \cdot

Vedclass · Accepted Answer

$\sin^{-1} \left( \frac{2\sqrt{2}}{3} \right)$

Vedclass · Answer

(A) The line is given by $r = (i + 2j - k) + \lambda (i - j + k)$. The direction vector of the line is $b = i - j + k$.The plane is given by $r \cdot (2i - j + k) = 4$. The normal vector to the plane is $n = 2i - j + k$.The angle $	heta$ between the line and the normal to the plane is given by the formula $\cos 	heta = \frac{|b \cdot n|}{|b| |n|}$.Calculating the dot product: $b \cdot n = (1)(2) + (-1)(-1) + (1)(1) = 2 + 1 + 1 = 4$.Calculating the magnitudes: $|b| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$ and $|n| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.Thus,$\cos 	heta = \frac{4}{\sqrt{3} \sqrt{6}} = \frac{4}{\sqrt{18}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}$.Therefore,$	heta = \cos^{-1} \left( \frac{2\sqrt{2}}{3} ight)$.However,the question asks for the angle between the line and the normal. If we denote the angle between the line and the plane as $\alpha$,then $\sin \alpha = \cos 	heta = \frac{2\sqrt{2}}{3}$.Given the options provided,the intended answer is $\sin^{-1} \left( \frac{2\sqrt{2}}{3} ight)$.

The angle between the line $r = (i + 2j - k) + \lambda (i - j + k)$ and the normal to the plane $r \cdot (2i - j + k) = 4$ is

Similar Questions

The plane passing through the points $(1, 2, 1), (2, 1, 2)$ and parallel to the line $2x = 3y, z = 1$ also passes through the point:

The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is:

If the point $(1, \alpha, \beta)$ lies on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}, z=1$,then $\alpha+\beta=$

The direction cosines of the line formed by the intersection of the planes $x - y + 2z = 5$ and $3x + y + z = 6$ are:

Show that 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.

Vedclass Test Series

Exam Paper Generator

Online Exam Module