Let vec{a} = 5hat{i} - hat{j} - 3hat{k} and v | vedclass

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

Question

(D) The projection of vector $\vec{a}$ on vector $\vec{b}$ is given by the formula: $	ext{Proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\ve

Vedclass · Accepted Answer

Projection of $\vec{a}$ on $\vec{b}$ is $\frac{-17}{\sqrt{35}}$ and the direction of the projection vector is opposite to the direction of $\vec{b}$

Vedclass · Answer

(D) The projection of vector $\vec{a}$ on vector $\vec{b}$ is given by the formula: $	ext{Proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.Given $\vec{a} = 5\hat{i} - \hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + 5\hat{k}$.First,calculate the dot product: $\vec{a} \cdot \vec{b} = (5)(1) + (-1)(3) + (-3)(5) = 5 - 3 - 15 = -13$.Next,calculate the magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{1^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.The scalar projection is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{-13}{\sqrt{35}}$.Since the scalar projection is negative,the direction of the projection vector is opposite to the direction of $\vec{b}$.Note: The provided options contain a calculation error in the numerator ($17$ instead of $13$). Based on the sign of the dot product,the correct choice is $D$.

Let $\vec{a} = 5\hat{i} - \hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + 5\hat{k}$ be two vectors. Then which one of the following statements is $TRUE$?

Similar Questions

The vectors $\vec{AB} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{BC} = \hat{i} - 2\hat{k}$ are the adjacent sides of a parallelogram. The angle between its diagonals is

In a $\triangle ABC$,let $G$ denote its centroid and let $M, N$ be points in the interiors of the segments $AB, AC$,respectively,such that $M, G, N$ are collinear. If $r$ denotes the ratio of the area of $\triangle AMN$ to the area of $\triangle ABC$,then

Let $|\vec{a}| = |\vec{b}| = 1$ and $|\vec{a} + \vec{b}| = \sqrt{3}$. If $\vec{c}$ is a vector satisfying the condition $\vec{c} - \vec{a} - 2\vec{b} = 3(\vec{a} \times \vec{b})$,then $\vec{c} \cdot \vec{b} = \dots$ (in $/2$)

In $\triangle ABC$,the points $P, Q, R$ divide $BC, CA, AB$ in the ratio $3:4, 2:5, 9:5$,respectively,and the point $D$ divides $BC$ in the ratio $2:3$. If $\vec{AP} + \vec{BQ} + \vec{CR} = k \vec{AD}$,then $(14k + 1) : (14k - 1) = $

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{b}$ is perpendicular to $\vec{c}$. If $|\vec{a}|=2, |\vec{b}|=3, |\vec{c}|=5$ and $|\vec{a}+\vec{b}+\vec{c}|=4 \sqrt{3}$,then the angle between $\vec{a}$ and $\vec{c}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module