Test Index

Vector Algebra Part 2

Section: Mathematics

© examsnet.com

Question : 86 of 100

Marks: +1, -0

Let

\vec{\alpha}=4\hat{i}+3\hat{j}+5\hat{k}

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

\vec{\beta}_1

\vec{\alpha}

\vec{\beta}_2

be perpendicular to

\vec{\alpha}

\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2

, then the value of

5\vec{\beta}_2\cdot(\hat{i}+\hat{j}+\hat{k})

[24-Jan-2023 Shift 2]

6
11
7
9

Solution:

Let

\vec{\beta}_1=\lambda\vec{\alpha}

Now

\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1

=(\hat{i}+2\hat{j}-4\hat{k})-\lambda(4\hat{i}+3\hat{j}+5\hat{k})

=(1-4\lambda)\hat{i}+(2-3\lambda)\hat{j}-(5\lambda+4)\hat{k}

\vec{\beta}_2\cdot\vec{\alpha}=0

\Rightarrow 4(1-4\lambda)+3(2-3\lambda)-5(5\lambda+4)=0

\Rightarrow 4-16\alpha+6-9\lambda-25\lambda-20=0

\Rightarrow 50\lambda=-10

\Rightarrow \lambda=\frac{-1}{5}

\vec{\beta}_2=\left(1+\frac{4}{5}\right)\hat{i}+\left(2+\frac{3}{5}\right)\hat{j}-(-1+4)\hat{k}

\vec{\beta}_2=\frac{9}{5}\hat{i}+\frac{13}{5}\hat{j}-3\hat{k}

5\vec{\beta}_2=9\hat{i}+13\hat{j}-15\hat{k}

5\vec{\beta}_2\cdot(\hat{i}+\hat{j}+\hat{k})=9+13-15=7

© examsnet.com

Go to Question:

Prev Question

Next Question