Consider the vectors vec{a}=3 hat{i}+5 hat{j} | vedclass

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

Question

(B) The length of the projection of vector $\vec{u}$ on vector $\vec{v}$ is given by $\frac{|\vec{u} \cdot \vec{v}|}{|\vec{v}|}$.$l = \frac{|\vec{a} \

Vedclass · Accepted Answer

$l=m=n$

Vedclass · Answer

(B) The length of the projection of vector $\vec{u}$ on vector $\vec{v}$ is given by $\frac{|\vec{u} \cdot \vec{v}|}{|\vec{v}|}$.$l = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} = \frac{|(3)(2) + (5)(-3) + (2)(-5)|}{\sqrt{2^2 + (-3)^2 + (-5)^2}} = \frac{|6 - 15 - 10|}{\sqrt{4 + 9 + 25}} = \frac{|-19|}{\sqrt{38}} = \frac{19}{\sqrt{38}}$.$m = \frac{|\vec{b} \cdot \vec{c}|}{|\vec{c}|} = \frac{|(2)(-5) + (-3)(-2) + (-5)(3)|}{\sqrt{(-5)^2 + (-2)^2 + 3^2}} = \frac{|-10 + 6 - 15|}{\sqrt{25 + 4 + 9}} = \frac{|-19|}{\sqrt{38}} = \frac{19}{\sqrt{38}}$.$n = \frac{|\vec{c} \cdot \vec{a}|}{|\vec{a}|} = \frac{|(-5)(3) + (-2)(5) + (3)(2)|}{\sqrt{3^2 + 5^2 + 2^2}} = \frac{|-15 - 10 + 6|}{\sqrt{9 + 25 + 4}} = \frac{|-19|}{\sqrt{38}} = \frac{19}{\sqrt{38}}$.Since $l = \frac{19}{\sqrt{38}}$, $m = \frac{19}{\sqrt{38}}$, and $n = \frac{19}{\sqrt{38}}$, we have $l = m = n$.

Consider the vectors $\vec{a}=3 \hat{i}+5 \hat{j}+2 \hat{k}$, $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$ and $\vec{c}=-5 \hat{i}-2 \hat{j}+3 \hat{k}$. If $l, m$ and $n$ are the lengths of the projections of $\vec{a}$ on $\vec{b}$, $\vec{b}$ on $\vec{c}$ and $\vec{c}$ on $\vec{a}$ respectively, then:

Similar Questions

In $\triangle OAB$, $O(0, 0, 0)$, $A(6, 2, -3)$ and $B(4, 0, 3)$ are the vertices. Let $\vec{a}$ and $\vec{b}$ be position vectors of points $A$ and $B$ respectively. If $OM$ is the projection of $\vec{a}$ on $\vec{b}$, then the length $l(AM)$ is equal to...

If $\vec{i}+\vec{j}-\vec{k}, -\vec{i}+2\vec{j}+\vec{k}, \vec{j}+2\vec{k}, 2\vec{i}-\vec{j}+2\vec{k}$ are the position vectors of four points $A, B, C, D$ respectively,then the shortest distance between the lines $AB$ and $CD$ is

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + \hat{k}$,and $\vec{c} = \hat{i} - \hat{j} - \hat{k}$ be three vectors. $A$ vector $\vec{v}$ in the plane of $\vec{a}$ and $\vec{b}$,whose projection on $\vec{c}$ is $1/\sqrt{3}$,is given by:

If $ABCDEF$ is a regular hexagon,the length of whose side is $a$,then $\overrightarrow{AB} \cdot \overrightarrow{AF} + \frac{1}{2} \overrightarrow{BC}^2 = $

If $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are the position vectors of the points $A, B, C, D$ respectively such that $3 \bar{a}-\bar{b}+2 \bar{c}-4 \bar{d}=\overline{0}$,then the position vector of the point of intersection of the line segments $AC$ and $BD$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module