A vector {a}=α{i}+{2j}+β{k} (α,β {R}) lies in the plane of the vectros {b}={i}+{j} and {c}={i}-{j}+{4k}. If {a} bisects the angle between {b} and {c}, then: [7-Jan-2020 Shift 1]

Correct Answer is : a⃗⋅k^+2=0{a}·{k}+2=0a⋅k^+2=0

Correct Answer is : a⃗⋅k^+2=0{a}·{k}+2=0a⋅k^+2=0

Test Index

Vector Algebra Part 1

Section: Mathematics

Question : 50 of 100

Marks: +1, -0

A vector

\vec{a}=\alpha\vec{i}+\vec{2j}+\beta\vec{k}

(\alpha,\beta \in \mathbb{R})

\vec{b}=\vec{i}+\vec{j}

\vec{c}=\vec{i}-\vec{j}+\vec{4k}

\vec{a}

\vec{b}

\vec{c}

Solution:

Angle bisector can be

\vec{a}=\lambda(\hat{b}+\hat{c}) \;\text{or}\; \vec{a}=u(\hat{b}-\hat{c})

\vec{a}=\lambda\left(\;\frac{\hat{i+\hat{j}}}{\sqrt{2}}+\;\frac{\hat{i-\hat{j}+4\hat{k}}}{3\sqrt{2}}\right)

\;\;=\;\frac{\lambda}{3\sqrt{2}}(3\hat{i}+3\hat{j}+\hat{i}-\hat{j}+4\hat{k})

\;\;=\;\frac{\lambda}{3\sqrt{2}}(4\hat{i}+2\hat{j}+4\hat{k}) \dots\dots\dots\dots\dots (i)

\;\text{also,}\; \vec{a}=\alpha\hat{i}+2\hat{j}+\beta\hat{k} \dots\dots\dots\dots\dots \text{ (ii)}\;

On comparing eq (1) and eq(2)

\;\frac{\lambda}{3\sqrt{2}}=1

\Rightarrow \lambda=3\sqrt{2}

putting value of

\lambda

in eq(i)

\vec{a}=4\hat{i}+2\hat{j}+4\hat{k}

\;\text{Now,}\; \vec{a}=u\left(\;\frac{\hat{i-\hat{j}}}{\sqrt{2}}-\;\frac{\hat{i-\hat{j}+4\hat{k}}}{3\sqrt{2}}\right)

=\;\frac{u}{3\sqrt{2}}(3\hat{i}+3\hat{j}-\hat{i}+\hat{j}-4\hat{k})

=\;\frac{u}{3\sqrt{2}}(2\hat{i}+4\hat{j}-4\hat{k})

=\;\frac{2u}{3\sqrt{2}}(\hat{i}+2\hat{j}-2\hat{k}) \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots (iii)

\;\text{on comparing eq(ii) and eq(iii)}\;

\;\frac{2u}{3\sqrt{2}}=1 \Rightarrow u=\;\frac{3\sqrt{2}}{2}

\;\text{putting value of}\; u \;\text{in eq(iii)}\;

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

\vec{a}\cdot \hat{k}=-2

\therefore \vec{a}\cdot\hat{k}+2=0

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