Let {a}={j}-{k} and {c}={i}-{j}-{k}. Then the vector {b} satisfying ({a} × {b})+{c}=0 and {a} · {b}=3, is

Correct Answer is : −i^+j^−2k^-{i}+{j}-2{k}−i^+j^−2k^

Correct Answer is : −i^+j^−2k^-{i}+{j}-2{k}−i^+j^−2k^

Test Index

UPSC NDA Math Model Paper 7

Question : 87 of 120

Marks: +1, -0

Let

\vec{a}=\hat{j}-\hat{k}

\vec{c}=\hat{i}-\hat{j}-\hat{k}

\vec{b}

(\vec{a} \times \vec{b})+\vec{c}=0

\vec{a} \cdot \vec{b}=3

, is

Solution:

Given,

\vec{a}=\hat{j}-\hat{k}

and

\vec{c}=\hat{i}-\hat{j}-\hat{k}

Consider,

(\vec{a} \times \vec{b})+\vec{c}=0

Multiply by

\vec{a}

on both side, we have

\Rightarrow \vec{a} \times [(\vec{a} \times \vec{b})+\vec{c}]=0

\Rightarrow \vec{a} \times [(\vec{a} \times \vec{b})]+(\vec{a} \times \vec{c})=0

\Rightarrow (\vec{a} \cdot \vec{b})\vec{a}-(\vec{a} \cdot \vec{a})\vec{b}+(\vec{a} \times \vec{c})=0

.... (1)

Given

\vec{a} \cdot \vec{b}=3, \vec{a} \cdot \vec{a}=2

and

\vec{a} \times \vec{b}=-2 \vec{i}-\vec{j}-\vec{k}

Equation 1!) becomes,

\Rightarrow 3 \vec{a}-2 \vec{b}+(-2 \vec{i}-\vec{j}-\vec{k})=0

\Rightarrow 2 \vec{b}=3 \vec{a}+(-2 \vec{i}-\vec{j}-\vec{k})

\Rightarrow 2 \vec{b}=3(\vec{j}-\vec{k})+(-2 \vec{i}-\vec{j}-\vec{k})

\Rightarrow \vec{b}=-\hat{i}+\hat{j}-2\hat{k}

Hence, if

\vec{a}=\hat{j}-\hat{k}

and

\vec{c}=\hat{i}-\hat{j}-\hat{k}

. Then the vector

\vec{b}

satisfying

(\vec{a} \times \vec{b})+\vec{c}=0

and

\vec{a} \cdot \vec{b}=3

, is

\vec{b}=-\hat{i}+\hat{j}-2\hat{k}

Go to Question:

Prev Question

Next Question