Question

The vectors $2\widehat{i}+3\widehat{j}-4\widehat{k}\ and\ a\widehat{i}+b\widehat{j}+c\widehat{k}$ are perpendicular, when
A) a=2, b=3, c=-4
B) a=4, b=4, c=5
C) a=4, b=4, c=-5
D) None of these

Answer 1

HINT: - Now, the fact that is important to know before answering this question is that the dot product of two vectors which are perpendicular is equal to 0 that can be written as follows
$\overrightarrow{v}\cdot \overrightarrow{u}=0$
(Where $\overrightarrow{v}\ and\ \overrightarrow{u}$ are two perpendicular vectors)
Another thing is that to calculate a dot product, we do the following
Let $\overrightarrow{v}=a\widehat{i}+b\widehat{j}+c\widehat{k}\ and\ \overrightarrow{u}=l\widehat{i}+m\widehat{j}+n\widehat{k}$ be two vectors of which we have to calculate the dot product.
Now, to calculate their dot product, we can write the following $$$$

& \Rightarrow \overrightarrow{v}\cdot \overrightarrow{u}=\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)\ \cdot \left( l\widehat{i}+m\widehat{j}+n\widehat{k} \right) \\\ & \Rightarrow \overrightarrow{v}\cdot \overrightarrow{u}=al+bm+cn \\\ \end{aligned}$$ _Complete step-by-step answer:_ As mentioned in the question, vectors $$2\widehat{i}+3\widehat{j}-4\widehat{k}\ and\ a\widehat{i}+b\widehat{j}+c\widehat{k}$$ are perpendicular to each other. Now, as we know that the dot product of two vectors which are perpendicular is equal to 0, hence, we can find the required values of a, b and c as follows $$\begin{aligned} & \Rightarrow \overrightarrow{v}\cdot \overrightarrow{u}=\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)\ \cdot \left( l\widehat{i}+m\widehat{j}+n\widehat{k} \right)=al+bm+cn \\\ & \Rightarrow \overrightarrow{v}\cdot \overrightarrow{u}=\left( 2\widehat{i}+3\widehat{j}-4\widehat{k} \right)\ \cdot \ \left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)=2a+3b-4c \\\ \end{aligned}$$ Now, the dot product must be zero, so we can substitute the values of a, b and c from the options and then we can get the right combination. Now, for (A), we get $$\begin{aligned} & \Rightarrow 2a+3b-4c \\\ & \Rightarrow 2\cdot 2+3\cdot 3-4\cdot (-4) \\\ & \Rightarrow 4+9+16 \\\ & \Rightarrow 29 \\\ \end{aligned}$$ As it is not equal to zero, hence, this is not the right combination. Now, for (B), we get $$\begin{aligned} & \Rightarrow 2a+3b-4c \\\ & \Rightarrow 2\cdot 4+3\cdot 4-4\cdot (5) \\\ & \Rightarrow 8+12-20 \\\ & \Rightarrow 0 \\\ \end{aligned}$$ As it is equal to zero, hence, this is the right combination and hence, the correct answer is (B). NOTE: An alternative method to solve this question is that we can take the cross product of the two vectors and as we know that the cross product of two perpendicular vectors is just the multiplication of the magnitudes of the two vectors and then, we can just put the values from the options and hence, find the correct answer.