The angle between the line bar{r}=(hat{i}+2ha | vedclass

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

Question

(C) The angle $	heta$ between the line $\bar{r}=\bar{a}+\lambda\bar{b}$ and the plane $\bar{r} \cdot \bar{n}=p$ is given by $\sin 	heta = \frac{|\ba

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The angle $	heta$ between the line $\bar{r}=\bar{a}+\lambda\bar{b}$ and the plane $\bar{r} \cdot \bar{n}=p$ is given by $\sin 	heta = \frac{|\bar{b} \cdot \bar{n}|}{|\bar{b}| |\bar{n}|}$.Here,$\bar{b} = \hat{i} - \hat{j} + \hat{k}$ and $\bar{n} = 2\hat{i} - \hat{j} + \hat{k}$.Calculating the dot product: $\bar{b} \cdot \bar{n} = (1)(2) + (-1)(-1) + (1)(1) = 2 + 1 + 1 = 4$.Calculating the magnitudes: $|\bar{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$ and $|\bar{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.Substituting these values into the formula: $\sin 	heta = \frac{4}{\sqrt{3} \cdot \sqrt{6}} = \frac{4}{\sqrt{18}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}$.Therefore,$	heta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}ight)$.

The angle between the line $\bar{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and the plane $\bar{r} \cdot (2\hat{i}-\hat{j}+\hat{k})=4$ is:

Similar Questions

Let $Q$ be the foot of the perpendicular drawn from the point $P(1, 2, 3)$ to the plane $x + 2y + z = 14$. If $R$ is a point on the plane such that $\angle PRQ = 60^{\circ}$,then the area of $\triangle PQR$ is equal to:

Let $P(3, 2, 6)$ be a point in space and $Q$ be a point on the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$. Then the value of $\mu$ for which the vector $\vec{PQ}$ is parallel to the plane $x - 4y + 3z = 1$ is

The equation of the plane containing the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and perpendicular to the plane containing the lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{1}$ and $\frac{x}{3}=\frac{y}{2}=\frac{z}{1}$ is

Let the equation of the plane,that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3x-2y+4z-7=0$ and $x+5y-2z+9=0$,be $\alpha x+\beta y+\gamma z+3=0$. Then $\alpha+\beta+\gamma$ is equal to:

The direction ratios of the normal to the plane passing through the points $(1, 2, -3)$, $(-1, -2, 1)$ and parallel to $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4}$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module