Question

If the vectors $\overrightarrow a = i - j + 2k,{\text{ }}\overrightarrow b = 2i + 4j + k,{\text{ }}\overrightarrow c = \lambda i + j + \mu k$ are mutually orthogonal, then $\left( {\lambda ,\mu } \right) = $
A) $\left( {2, - 3} \right)$
B) $\left( { - 2,3} \right)$
C) $\left( {3, - 2} \right)$
D) $\left( { - 3,2} \right)$

Answer 1

Hint: As we are given $\overrightarrow a ,{\text{ }}\overrightarrow b ,{\text{ }}\overrightarrow c $ are mutually perpendicular, so , if the two vectors are perpendicular, then, $\overrightarrow c .\overrightarrow b = 0{\text{ and }}\overrightarrow a. \overrightarrow c = 0$. Both dot products will be zero.

Complete step-by-step answer:
Three vectors are given, $\overrightarrow a ,{\text{ }}\overrightarrow b ,{\text{ }}\overrightarrow c $ and it is said that all three are mutually perpendicular. That means the vectors are perpendicular to each other, that is, $\overrightarrow a \bot \overrightarrow b ,{\text{ }}\overrightarrow b \bot \overrightarrow c ,{\text{ }}\overrightarrow a \bot \overrightarrow c $. $ \bot $ is the sign of perpendicular.
Now we are provided that
$
  \overrightarrow a = i - j + 2k \\
  {\text{ }}\overrightarrow b = 2i + 4j + k \\
  \overrightarrow c = \lambda i + j + \mu k \\
 $
As they are mutually perpendicular, so their dot product is zero.
So, $\overrightarrow c .\overrightarrow b = 0{\text{ and }}\overrightarrow a .\overrightarrow c = 0$.
Now first let's do $\overrightarrow b .\overrightarrow c = 0$
$
  \left( {2i + 4j + k} \right).\left( {\lambda i + j + \mu k} \right) = 0 \\
  2\lambda + 4 + \mu = 0 \\
 $
$2\lambda + \mu = - 4$ (1)
Now upon solving $\overrightarrow a .\overrightarrow c = 0$
$
  \left( {i - j + 2k} \right).\left( {\lambda i + j + \mu k} \right) = 0 \\
  \lambda - 1 + 2\mu = 0 \\
 $
$\lambda + 2\mu = 1$ (2)
Now from equation (1), we get
$\mu = - 2\lambda - 4$
Putting this value in equation (2)
$
  \lambda + 2\left( { - 2\lambda - 4} \right) = 1 \\
  \lambda - 4\lambda - 8 = 1 \\
   - 3\lambda = 9 \\
  \lambda = - 3 \\
  {\text{And}} \\
  \mu = - 2\lambda - 4 \\
   = - 2 \times \left( { - 3} \right) - 4 \\
  \mu = 2 \\
 $
So we get $\left( {\lambda ,\mu } \right) = \left( { - 3,2} \right)$

Hence option D is correct.

Note: We know when $\overrightarrow a {\text{ and }}\overrightarrow b $ are perpendicular, then $\overrightarrow a. \overrightarrow b = 0$. Or when $\overrightarrow a {\text{ and }}\overrightarrow b $ are parallel, then $\overrightarrow a \times \overrightarrow b = 0$. As we know, $\overrightarrow a .\overrightarrow b = ab\cos \theta $ and if they are perpendicular, then the angle between them or $\theta = {90^ \circ }$. So, $ab\cos {90^ \circ } = 0$. Similarly, we know that $\overrightarrow a \times \overrightarrow b = ab\sin \theta $. It is zero when the angle between them is $0$. So they are parallel.