Q82 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
Find the value of '$b$' such that the scalar product of the vector $\hat{\imath}+\hat{\jmath}+\hat{k}$ with the unit vector parallel to the sum of the vectors $2 \hat{\imath}+4 \hat{\jmath}-5 \hat{k}$ and $b \hat{\imath}+2 \hat{\jmath}+3 \hat{k}$ is unity
- Correct answerA1
- B0
- C$-1$
- D$-2$
Solution
Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$.
Let $\vec{u} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{v} = b\hat{i} + 2\hat{j} + 3\hat{k}$.
The sum of the vectors is $\vec{s} = \vec{u} + \vec{v} = (2+b)\hat{i} + 6\hat{j} - 2\hat{k}$.
The unit vector parallel to $\vec{s}$ is $\hat{s} = \dfrac{\vec{s}}{|\vec{s}|} = \dfrac{(2+b)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2+b)^2 + 6^2 + (-2)^2}}$.
The scalar product of $\vec{a}$ and $\hat{s}$ is given as unity:
$\vec{a} \cdot \hat{s} = 1 \Rightarrow \dfrac{(1)(2+b) + (1)(6) + (1)(-2)}{\sqrt{(2+b)^2 + 36 + 4}} = 1$.
Simplifying the numerator: $2 + b + 6 - 2 = b + 6$.
So, $b + 6 = \sqrt{(2+b)^2 + 40}$.
Squaring both sides: $(b+6)^2 = (2+b)^2 + 40$.
$b^2 + 12b + 36 = b^2 + 4b + 4 + 40$.
$12b + 36 = 4b + 44$.
$8b = 8$.
$b = 1$.
Answer: 1
Let $\vec{u} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{v} = b\hat{i} + 2\hat{j} + 3\hat{k}$.
The sum of the vectors is $\vec{s} = \vec{u} + \vec{v} = (2+b)\hat{i} + 6\hat{j} - 2\hat{k}$.
The unit vector parallel to $\vec{s}$ is $\hat{s} = \dfrac{\vec{s}}{|\vec{s}|} = \dfrac{(2+b)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2+b)^2 + 6^2 + (-2)^2}}$.
The scalar product of $\vec{a}$ and $\hat{s}$ is given as unity:
$\vec{a} \cdot \hat{s} = 1 \Rightarrow \dfrac{(1)(2+b) + (1)(6) + (1)(-2)}{\sqrt{(2+b)^2 + 36 + 4}} = 1$.
Simplifying the numerator: $2 + b + 6 - 2 = b + 6$.
So, $b + 6 = \sqrt{(2+b)^2 + 40}$.
Squaring both sides: $(b+6)^2 = (2+b)^2 + 40$.
$b^2 + 12b + 36 = b^2 + 4b + 4 + 40$.
$12b + 36 = 4b + 44$.
$8b = 8$.
$b = 1$.
Answer: 1
Hence, the correct answer is option: 'A'.