If O(0,0,0),A(1,2,1),B(2,1,3),and C(-1,1,2) a | vedclass

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

Question

(B) $1$. Find the normal vector $\vec{n}$ to the face $OAB$. The vectors $\vec{OA} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{OB} = 2\hat{i} + \hat{j}

Vedclass · Accepted Answer

$\operatorname{Sin}^{-1}\left(\frac{6 \sqrt{2}}{5 \sqrt{7}}\right)$

Vedclass · Answer

(B) $1$. Find the normal vector $\vec{n}$ to the face $OAB$. The vectors $\vec{OA} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{OB} = 2\hat{i} + \hat{j} + 3\hat{k}$ lie on the face $OAB$. $2$. The normal vector $\vec{n} = \vec{OA} 	imes \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3-2) + \hat{k}(1-4) = 5\hat{i} - \hat{j} - 3\hat{k}$. $3$. The vector representing edge $BC$ is $\vec{BC} = \vec{OC} - \vec{OB} = (-1-2)\hat{i} + (1-1)\hat{j} + (2-3)\hat{k} = -3\hat{i} + 0\hat{j} - \hat{k}$. $4$. The angle $	heta$ between a line with direction vector $\vec{v}$ and a plane with normal $\vec{n}$ is given by $\sin 	heta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$. $5$. $\vec{v} \cdot \vec{n} = (-3)(5) + (0)(-1) + (-1)(-3) = -15 + 0 + 3 = -12$. $6$. $|\vec{v}| = \sqrt{(-3)^2 + 0^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10}$. $7$. $|\vec{n}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25+1+9} = \sqrt{35}$. $8$. $\sin 	heta = \frac{|-12|}{\sqrt{10} \sqrt{35}} = \frac{12}{\sqrt{350}} = \frac{12}{5\sqrt{14}} = \frac{12}{5\sqrt{2}\sqrt{7}} = \frac{6\sqrt{2}}{5\sqrt{7}}$. $9$. Thus,$	heta = \operatorname{Sin}^{-1}\left(\frac{6\sqrt{2}}{5\sqrt{7}}ight)$.

If $O(0,0,0)$,$A(1,2,1)$,$B(2,1,3)$,and $C(-1,1,2)$ are the vertices of a tetrahedron,then the acute angle between its face $OAB$ and edge $BC$ is

Similar Questions

The point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane $2x + 3y + z = 0$ is

Show that the points $(\hat{i}-\hat{j}+3 \hat{k})$ and $3(\hat{i}+\hat{j}+\hat{k})$ are equidistant from the plane $\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0$ and lie on opposite sides of it.

The equation of a plane at a distance of $\sqrt{\frac{2}{21}}$ from the origin,which contains the line of intersection of the planes $x-y-z-1=0$ and $2x+y-3z+4=0$,is:

$L$ is a line passing through the point $A(1, 0, -3)$ and parallel to a line having direction ratios $0, 1, -2$. $P$ is a point on the line $L$ which is at a minimum distance from the plane $2x + 3y + 5z = 1$. Then, the equation of the plane through $P$ and perpendicular to $AP$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module