Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Vectors

Question:

If $|\vec{a}+\vec{b}|=15, |\vec{a}-\vec{b}|=10, |\vec{a}|=\frac{11}{2}$ then value of $|\vec{b}|$ is/are :

Options:

$\frac{23}{2}$

$±\frac{23}{2}$

$±23$

$\frac{23}{\sqrt{2}}$

Correct Answer:

$\frac{23}{2}$

Explanation:

The correct answer is Option (1) → $\frac{23}{2}$

$|(\vec{a}+\vec{b}).(\vec{a}-\vec{b})|=|a^2-b^2|$

$(\vec{a}+\vec{b}).(\vec{a}-\vec{b})=|a^2-b^2|^2$

$⇒|\vec{a}|^2+|\vec{b}|^2+2\vec{a}.\vec{b}=|a^2+b^2|^2$  ...(1)

similarly $|\vec{a}|^2+|\vec{b}|^2-2\vec{a}.\vec{b}=|a^2-b^2|^2$  ...(2)

eq. (1) + eq. (2)

$⇒2(|\vec{a}|^2+|\vec{b}|^2)=|a^2+b^2|^2+|a^2-b^2|^2$

$2(\frac{121}{4}+|\vec{b}|^2)=225+100$

$⇒|\vec{b}|^2=\frac{529}{4}⇒|\vec{b}|=\frac{23}{2}$