Find the angle between the line frac{x+1}{2}= | vedclass

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

Question

(A) The direction vector of the line is $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.The normal vector to the plane is $\vec{n} = 10\hat{i} + 2\hat{j} -

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{8}{21}\right)$

Vedclass · Answer

(A) The direction vector of the line is $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.The normal vector to the plane is $\vec{n} = 10\hat{i} + 2\hat{j} - 11\hat{k}$.Let $\phi$ be the angle between the line and the plane. The formula for the angle is given by $\sin \phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.First,calculate the dot product: $\vec{b} \cdot \vec{n} = (2)(10) + (3)(2) + (6)(-11) = 20 + 6 - 66 = -40$.Next,calculate the magnitudes: $|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.$|\vec{n}| = \sqrt{10^2 + 2^2 + (-11)^2} = \sqrt{100 + 4 + 121} = \sqrt{225} = 15$.Thus,$\sin \phi = \frac{|-40|}{7 	imes 15} = \frac{40}{105} = \frac{8}{21}$.Therefore,$\phi = \sin^{-1}\left(\frac{8}{21}ight)$.

Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y-11z=3$.

Similar Questions

Find the equation of the plane containing the lines $\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - 2\hat{k})$.

Let $L$ be a line obtained from the intersection of two planes $x+2y+z=6$ and $y+2z=4$. If point $P(\alpha, \beta, \gamma)$ is the foot of the perpendicular from $(3,2,1)$ on $L$,then the value of $21(\alpha+\beta+\gamma)$ equals ...... .

The foot of the perpendicular drawn from $A(1, 2, 2)$ onto the plane $x+2y+2z-5=0$ is $B(\alpha, \beta, \gamma)$. If $\pi(x, y, z) \equiv x+2y+2z+5=0$ is a plane, then $-\pi(A) : \pi(B) =$ ?

The point of intersection of the line $x+1=\frac{y+3}{3}=\frac{-z+2}{2}$ with the plane $3x+4y+5z=10$ is

The plane passing through the line $L: \ell x-y+3(1-\ell)z=1, x+2y-z=2$ and perpendicular to the plane $3x+2y+z=6$ is $3x-8y+7z=4$. If $\theta$ is the acute angle between the line $L$ and the $y$-axis,then $415 \cos^{2} \theta$ is equal to...

Vedclass Test Series

Exam Paper Generator

Online Exam Module