MCQ
Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}, \vec{a} \cdot \vec{b}=3 \quad$ and $\quad|\vec{a} \times \vec{b}|^{2}=75$.Then $|\vec{a}|^{2}$ is equal to $.......$
  • $14$
  • B
    $13$
  • C
    $12$
  • D
    $11$

Answer

Correct option: A.
$14$
a
$|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2} ; \vec{a} \cdot \vec{b}=3$

As $|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+2|\vec{b}|^{2}$

$|\vec{b}|^{2}=2 \vec{a} \cdot \vec{b}=6$

$|\vec{a} \times \vec{b}|^{2}=75$

$|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}=75$

$6|\vec{a}|^{2}-9=75 \Rightarrow|\vec{a}|^{2}=14$

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