If vec{a}=hat{i}+2 hat{j}-3 hat{k} and vec{b} | vedclass

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

Question

(B) Given $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$.First,calculate $\vec{a}-\vec{b} = (1-2)\hat{i} + (2-(-3))

Vedclass · Accepted Answer

$|\vec{a}-\vec{b}|>|\vec{b}|-|\vec{a}|$

Vedclass · Answer

(B) Given $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$.First,calculate $\vec{a}-\vec{b} = (1-2)\hat{i} + (2-(-3))\hat{j} + (-3-(-5))\hat{k} = -\hat{i} + 5\hat{j} + 2\hat{k}$.The magnitude $|\vec{a}-\vec{b}| = \sqrt{(-1)^2 + 5^2 + 2^2} = \sqrt{1+25+4} = \sqrt{30}$.The magnitude $|\vec{a}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1+4+9} = \sqrt{14}$.The magnitude $|\vec{b}| = \sqrt{2^2 + (-3)^2 + (-5)^2} = \sqrt{4+9+25} = \sqrt{38}$.Now,check the options. For option $B$,$|\vec{b}|-|\vec{a}| = \sqrt{38} - \sqrt{14} \approx 6.16 - 3.74 = 2.42$.Since $|\vec{a}-\vec{b}| = \sqrt{30} \approx 5.47$,we have $5.47 > 2.42$.Thus,$|\vec{a}-\vec{b}| > |\vec{b}|-|\vec{a}|$ is correct.

If $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$,then

Similar Questions

In a quadrilateral $ABCD$,$M$ and $N$ are the mid-points of the sides $AB$ and $CD$ respectively. If $\vec{AD} + \vec{BC} = t \vec{MN}$,then $t =$

The unit vector in the direction of the vector $(1, 0, 0)$ is $.......$

$A$ zero vector has

The value of $k$ for which the vectors $\vec{a} = \hat{i} - \hat{j}$ and $\vec{b} = -2\hat{i} + k\hat{j}$ are collinear is

Vedclass Test Series

Exam Paper Generator

Online Exam Module