Q. If the scalar product of the vector $\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}\$ with the unit vector along $m\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}$ is equal to 2, then one of the value of $m$ is

A) 3
B) 4
C) 5
D) 6
Explanation

(d)
$\left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right) \bullet \left( \frac{m\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}}{\sqrt{13 + m^{2}}} \right) = 2$
$\Longrightarrow m + 2 + 6 = 2\sqrt{13 + m^{2}}$
$\Longrightarrow (m + 8)^{2} = 4(13 + m^{2})$
$\Longrightarrow m^{2} + 16m + 64 = 4m^{2} + 52$
$\Longrightarrow {3m}^{2} - 16m - 12 = 0$
$\Longrightarrow (3m + 2)(m - 6) = 0$
$\Longrightarrow m = 6, - \frac{2}{3}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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