Vector Algebra Question 2
Question 2 - 24 January - Shift 2
Let $\vec{\alpha}=4 i+3 j+5 k$ and $\beta=i+2 j-4 k$. Let $\beta_1$ be parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ be perpendicular to $\vec{\alpha}$. If $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$, then the value of $5 \vec{\beta}_2 \cdot(\hat{i}+\hat{j}+\hat{k})$ is
(1) 6
(2) 11
(3) 7
(4) 9
Show Answer
Answer: (3)
Solution:
Formula: Scalar product of two vectors when two vectors are parallel or perpendicular.
Let $\vec{\beta}_1=\lambda \vec{\alpha}$
Now $\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1$
$=(\hat{i}+2 \hat{j}-4 \hat{k})-\lambda(4 \hat{i}+3 \hat{j}+5 \hat{k})$
$=(1-4 \lambda) \hat{i}+(2-3 \lambda) \hat{j}-(5 \lambda+4) \hat{k}$
$\vec{\beta}_2 \cdot \vec{\alpha}=0$
$\Rightarrow 4(1-4 \lambda)+3(2-3 \lambda)-5(5 \lambda+4)=0$
$\Rightarrow 4-16 \alpha+6-9 \lambda-25 \lambda-20=0$
$\Rightarrow 50 \lambda=-10$
$\Rightarrow \lambda=\frac{-1}{5}$
$\vec{\beta}_2=(1+\frac{4}{5}) \hat{i}+(2+\frac{3}{5}) \hat{j}-(-1+4) \hat{k}$
$\vec{\beta}_2=\frac{9}{5} \hat{i}+\frac{13}{5} \hat{j}-3 \hat{k}$
$5 \vec{\beta}_2=9 \hat{i}+13 \hat{j}-15 \hat{k}$
$5 \vec{\beta}_2 \cdot(\hat{i}+\hat{j}+\hat{k})=9+13-15=7$
