Question

The vectors $2\hat i + 3\hat j - 4\hat k$ and $a\hat i + b\hat j + c\hat k$ are perpendicular, when
A. \[a = 2,{\text{ }}b = 3,{\text{ }}c = - 4\]
B. \[a = 4,{\text{ }}b = 4,{\text{ }}c = 5\]
C. \[a = 4,{\text{ }}b = 4{\text{ }}c = - 5\]
D. none of these

Answer 1

Hint: We know that the dot product of two perpendicular vectors is equal to zero, using this we’ll form an equation in a, b, c. Then we’ll substitute the values from the option to that equation to check which of the options are correct and hence will acquire the answer.

Complete step by step Answer:

Given data: two vectors $2\hat i + 3\hat j - 4\hat k$ and $a\hat i + b\hat j + c\hat k$
We know that the dot product of two perpendicular vectors is equal to zero
Therefore two vectors $2\hat i + 3\hat j - 4\hat k$ and $a\hat i + b\hat j + c\hat k$ will perpendicular if their dot product is zero
i.e. $(2\hat i + 3\hat j - 4\hat k).(a\hat i + b\hat j + c\hat k) = 0$
using $\hat i.\hat i = \hat j.\hat j{\text{ = }}\hat k.\hat k = 1$, $\hat i.\hat j = \hat j.\hat i = 0$, $\hat i.\hat k = \hat k.\hat i = 0$and $\hat j.\hat k{\text{ = }}\hat k.\hat j = 0$
$ \Rightarrow 2a + 3b - 4c = 0...........(i)$
Now, substituting the values of option in equation(i)
Option(i) \[a = 2,{\text{ }}b = 3,{\text{ }}c = - 4\]
Substituting values in the left-hand side
$ \Rightarrow 2(2) + 3(3) - 4( - 4)$
Since, $2(2) + 3(3) - 4( - 4) \ne 0$
It is not the correct option
Option(ii) \[a = 2,{\text{ }}b = 3,{\text{ }}c = - 4\]
Substituting values in the left-hand side
$ \Rightarrow 2(4) + 3(4) - 4(5)$
$ \Rightarrow 8 + 12 - 20$
$ \Rightarrow 0$
Since, $2(a) + 3(b) - 4(c) = 0$
It is the correct option
Option(B) is correct
Option(iii) \[a = 4,{\text{ }}b = 4,{\text{ }}c = - 5\]
Substituting values in the left-hand side
$ \Rightarrow 2(4) + 3(4) - 4( - 5)$
Since, $2(4) + 3(4) - 4( - 5) \ne 0$
It is not the correct option.
Therefore, from the above results, we can say that option(B) is the correct option

Note: In the above solution we mentioned that the dot product of two perpendicular vectors is zero, the actual formula for the dot product of two vectors is given by
$\vec a.\vec b = \left| {\vec a} \right|\left| {\vec b} \right|\cos \theta $, where $\theta $ is the angle between the two vectors. Now, for perpendicular vectors $\theta = {90^ \circ }$i.e. $\cos \theta = 0$, hence the dot product results in zero.