If P=(0,1,0) and Q=(0,0,1),then the length of | vedclass

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

Question

(B) The vector representing the line segment $PQ$ is $\vec{PQ} = (0-0, 0-1, 1-0) = (0, -1, 1)$.The normal vector to the plane $x+y+z=3$ is $\vec{n} =

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(B) The vector representing the line segment $PQ$ is $\vec{PQ} = (0-0, 0-1, 1-0) = (0, -1, 1)$.The normal vector to the plane $x+y+z=3$ is $\vec{n} = (1, 1, 1)$.The length of the projection of a vector $\vec{v}$ onto a plane with normal $\vec{n}$ is given by the formula $L = |\vec{v}| \sin(	heta)$,where $	heta$ is the angle between the vector $\vec{v}$ and the normal $\vec{n}$.First,calculate the magnitude of $\vec{PQ}$:$|\vec{PQ}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.Next,calculate the cosine of the angle $	heta$ between $\vec{PQ}$ and $\vec{n}$:$\cos(	heta) = \frac{|\vec{PQ} \cdot \vec{n}|}{|\vec{PQ}| |\vec{n}|} = \frac{|(0)(1) + (-1)(1) + (1)(1)|}{\sqrt{2} \cdot \sqrt{1^2+1^2+1^2}} = \frac{|0 - 1 + 1|}{\sqrt{2} \cdot \sqrt{3}} = 0$.Since $\cos(	heta) = 0$,the angle $	heta = 90^\circ$,which means $\sin(	heta) = 1$.Therefore,the length of the projection is $L = |\vec{PQ}| \sin(90^\circ) = \sqrt{2} \cdot 1 = \sqrt{2}$.

If $P=(0,1,0)$ and $Q=(0,0,1)$,then the length of the projection of the line segment $PQ$ on the plane $x+y+z=3$ is:

Similar Questions

The point where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}$ meets the plane $2x+4y-z=1$ is:

The sine of the angle between the straight line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ and the plane $2x-2y+z=5$ is

The angle between the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 3}{-2}$ and the plane $x + y + 4 = 0$ is ......... $^o$.

If the line $\bar{r}=(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}+\hat{\jmath}+2 \hat{k})$ is parallel to the plane $\bar{r} \cdot (3 \hat{\imath}-2 \hat{\jmath}+m \hat{k})=10$,then the value of $m$ is

Let $L$ be the line of intersection of the planes $2x+3y+z=1$ and $x+3y+2z=1$. If $L$ makes an angle $\alpha$ with the positive $x$-axis,then $\cos \alpha$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module