Let the vectors overrightarrow{a}, overrighta | vedclass

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

Question

(A) Given that the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is equal to the projection of $\overrightarrow{c}$ on $\overrightarrow{a

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Given that the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is equal to the projection of $\overrightarrow{c}$ on $\overrightarrow{a}$.$\Rightarrow \frac{\overrightarrow{b} \cdot \overrightarrow{a}}{|\overrightarrow{a}|} = \frac{\overrightarrow{c} \cdot \overrightarrow{a}}{|\overrightarrow{a}|} \Rightarrow \overrightarrow{b} \cdot \overrightarrow{a} = \overrightarrow{c} \cdot \overrightarrow{a}$.Also,$\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$,so $\overrightarrow{b} \cdot \overrightarrow{c} = 0$.Let $k = |\overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c}|$.Squaring both sides,we get:$k^2 = |\overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b}) - 2(\overrightarrow{a} \cdot \overrightarrow{c}) - 2(\overrightarrow{b} \cdot \overrightarrow{c})$.Since $\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{a} \cdot \overrightarrow{c}$ and $\overrightarrow{b} \cdot \overrightarrow{c} = 0$,the terms cancel out:$k^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{a} \cdot \overrightarrow{b} - \overrightarrow{a} \cdot \overrightarrow{c}) - 2(0) = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2$.Substituting the given magnitudes $|\overrightarrow{a}|=2, |\overrightarrow{b}|=4, |\overrightarrow{c}|=4$:$k^2 = 2^2 + 4^2 + 4^2 = 4 + 16 + 16 = 36$.Therefore,$k = \sqrt{36} = 6$.

Similar Questions

If vector $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$ and vector $\vec{b} = -2\hat{i} + 2\hat{j} - \hat{k},$ then $\frac{\text{Projection of vector } \vec{a} \text{ on vector } \vec{b}}{\text{Projection of vector } \vec{b} \text{ on vector } \vec{a}} = $

$A$ parallelogram is constructed with $5\vec{a} + 2\vec{b}$ and $\vec{a} - 3\vec{b}$ as its adjacent sides, where $|\vec{a}| = 2\sqrt{2}$ and $|\vec{b}| = 3$. The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$. Find the lengths of the diagonals of the parallelogram.

If $x$ and $y$ are two unit vectors and $\theta$ is the angle between them,then $\frac{1}{2}|x-y|$ is equal to

Find the projection of the vector $\hat{i}+3 \hat{j}+7 \hat{k}$ on the vector $7 \hat{i}-\hat{j}+8 \hat{k}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module