The magnitude of the projection of the vector | vedclass

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

Question

(B) Let $\vec{a} = 2\hat{i}+\hat{j}+\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,and $\vec{c} = \hat{i}+2\hat{j}+3\hat{k}$.The vector perpendicular to

Vedclass · Accepted Answer

$\frac{1}{\sqrt{6}}$ units

Vedclass · Answer

(B) Let $\vec{a} = 2\hat{i}+\hat{j}+\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,and $\vec{c} = \hat{i}+2\hat{j}+3\hat{k}$.The vector perpendicular to the plane containing $\vec{b}$ and $\vec{c}$ is given by $\vec{n} = \vec{b} 	imes \vec{c}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1) = \hat{i} - 2\hat{j} + \hat{k}$.The magnitude of the projection of $\vec{a}$ on $\vec{n}$ is given by $\left| \frac{\vec{a} \cdot \vec{n}}{|\vec{n}|} ight|$.$\vec{a} \cdot \vec{n} = (2)(1) + (1)(-2) + (1)(1) = 2 - 2 + 1 = 1$.$|\vec{n}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.Therefore,the required projection is $\frac{1}{\sqrt{6}}$ units.

The magnitude of the projection of the vector $2\hat{i}+\hat{j}+\hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+3\hat{k}$ is:

Similar Questions

$A$ tetrahedron has vertices $P(1, 2, 1)$,$Q(2, 1, 3)$,$R(-1, 1, 2)$ and $O(0, 0, 0)$. The angle between the faces $OPQ$ and $PQR$ is

The area of the triangle whose vertices are $A(1, 1, 1)$,$B(1, 2, 3)$,and $C(2, 3, 1)$ is . . . . . . .

Vectors $\bar{a}$ and $\bar{b}$ are such that $|\bar{a}|=1$,$|\bar{b}|=4$ and $\bar{a} \cdot \bar{b}=2$. If $\bar{c}=2 \bar{a} \times \bar{b}-3 \bar{b}$,then the angle between $\bar{b}$ and $\bar{c}$ is

If $a+2b+3c=0$ and $(a \times b)+(b \times c)+(c \times a)=\lambda(b \times c)$,then the value of $\lambda$ is equal to

Let $\vec{u}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{v}=-3 \hat{j}+2 \hat{k}$ be vectors in $R^3$ and $\vec{w}$ be a unit vector in the $XY$-plane. Then,the maximum value of $|(\vec{u} \times \vec{v}) \cdot \vec{w}|$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module