A vector perpendicular to the plane containin | vedclass

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

Question

(B) To find a vector perpendicular to the plane containing points $A, B$,and $C$,we calculate the cross product of two vectors lying in the plane,such

Vedclass · Accepted Answer

$8 \hat{i} + 4 \hat{j} + 4 \hat{k}$

Vedclass · Answer

(B) To find a vector perpendicular to the plane containing points $A, B$,and $C$,we calculate the cross product of two vectors lying in the plane,such as $\vec{AB}$ and $\vec{AC}$.Given points: $A(1, -1, 2)$,$B(2, 0, -1)$,$C(0, 2, 1)$.$\vec{AB} = (2-1)\hat{i} + (0 - (-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + \hat{j} - 3\hat{k}$$\vec{AC} = (0-1)\hat{i} + (2 - (-1))\hat{j} + (1-2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}$The normal vector $\vec{n} = \vec{AB} 	imes \vec{AC} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -3 \ -1 & 3 & -1 \end{array}ight|$$\vec{n} = \hat{i}(-1 - (-9)) - \hat{j}(-1 - 3) + \hat{k}(3 - (-1))$$\vec{n} = \hat{i}(8) - \hat{j}(-4) + \hat{k}(4) = 8\hat{i} + 4\hat{j} + 4\hat{k}$Thus,the vector perpendicular to the plane is $8\hat{i} + 4\hat{j} + 4\hat{k}$.

$A$ vector perpendicular to the plane containing the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ is

Similar Questions

$i \times (j \times k) = $

If non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular to each other,then the solution of the equation $\vec{r} \times \vec{a} = \vec{b}$ is:

If $\bar{a}, \bar{b}, \bar{c}$ are three coplanar vectors such that $|\bar{a}|=1, |\bar{b}|=2$,$\bar{b} \cdot \bar{c}=8$,and the angle between $\bar{b}$ and $\bar{c}$ is $45^{\circ}$,then $|\bar{a} \times (\bar{b} \times \bar{c})|=$

Let $\vec{a}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+3 \hat{k}$. If $\vec{b}$ is a vector such that $\vec{a}=\vec{b} \times \vec{c}$ and $|\vec{b}|^2=50$,then $|72-| \vec{b}+\vec{c}|^2 |$ is equal to $..........$.

$\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=\hat{i}+2 \hat{j}-\hat{k}$ are two vectors and $\vec{a}$ is a unit vector such that $\cos (\vec{a}, \vec{b} \times \vec{c})=\sqrt{\frac{2}{3}}$. Then $|\vec{a} \times(\vec{b} \times \vec{c})|=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module