The direction cosines of the normal to the pl | vedclass

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

Question

(B) The equation of the plane is given by $2x + 3y - 6z = 5$.Comparing this with the general equation of a plane $Ax + By + Cz = D$,the direction rati

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The equation of the plane is given by $2x + 3y - 6z = 5$.Comparing this with the general equation of a plane $Ax + By + Cz = D$,the direction ratios of the normal to the plane are $(A, B, C) = (2, 3, -6)$.To find the direction cosines $(l, m, n)$,we divide the direction ratios by the magnitude of the normal vector,which is $\sqrt{A^2 + B^2 + C^2}$.Magnitude $= \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.Therefore,the direction cosines are $\left( \frac{2}{7}, \frac{3}{7}, -\frac{6}{7} \right)$.

The direction cosines of the normal to the plane $2x + 3y - 6z = 5$ are

Similar Questions

$A$ line makes the same angle $\theta$ with each of the $x$ and $z$ axes. If the angle $\beta$ which it makes with the $y$-axis is such that $\sin^2 \beta = 3\sin^2 \theta$,then $\cos^2 \theta$ is-

Fill in the blanks:
The coordinates of points in the $XY$-plane are of the form ...........

The angle between the lines whose direction ratios are proportional to $(1, 2, 1)$ and $(2, -3, 6)$ is

Each of the angles $\beta$ and $\gamma$ that a given line makes with the positive $y-$ and $z-$axes,respectively,is half of the angle that this line makes with the positive $x-$axis. Then the sum of all possible values of the angle $\beta$ is

If the direction cosines of two lines are such that $l+m+n=0$ and $l^2+m^2-n^2=0$, then the angle between them is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module