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Vector Algebra question

2022 · 29 Jul · Shift 2 · Q42

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JEE Main
Mathematics
Vector Algebra
2022 · 29 Jul · Shift 2 · Q42

Vector Algebra question

2022 · 29 Jul · Shift 2 · Q42

JEE MainMathematicsVector AlgebraNumerical+4 / −1

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

and

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

. Then

|\vec{a}|^{2}

is equal to

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 14

Let $|vec a|^2=x,qquad |vec b|^2=y.$ We are given: $|vec a+vec b|^2=|vec a|^2+2|vec b|^2.$
Use the identity

dot vec b.$$ So, $$x+y+2(vec a dot vec b)=x+2y.$$ Given $$vec a dot vec b=3,$$ we get $$x+y+2(3)=x+2y$$ $$y+6=2y$$ $$y=6.$$ Thus, $$|vec b|^2=6.$$ 3. Now use the identity for cross product: $$|vec a imes vec b|^2=|vec a|^2|vec b|^2-(vec a dot vec b)^2.$$ Given $$|vec a imes vec b|^2=75,$$ so $$xy-3^2=75.$$ Substitute $y=6$: $$6x-9=75$$ $$6x=84$$ $$x=14.$$ 4. Therefore, $$|vec a|^2=14.$$

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