Question

If a vector $2\widehat{i}+3\widehat{j}+8\widehat{k}$ is perpendicular to the vector $4\widehat{i}-4\widehat{j}+a\widehat{k}$, then the value of a is –

& \text{A) }-1 \\\ & \text{B) }\dfrac{1}{2} \\\ & \text{C) }-\dfrac{1}{2} \\\ & \text{D) 1} \\\ \end{aligned}$$

Answer 1

Two vectors related to each other as being parallel or perpendicular follows many conditions. We can employ the different conditions that give us favourable outcomes to find the unknown coefficient in the vector form of the second equation.

Complete answer:
We know that the vectors can be multiplied in two different ways – The vector product or the cross product and the scalar product or the dot product. The former gives a vector which has direction perpendicular to the given vectors and the latter product gives a scalar quantity without any direction sense.
Let us consider our problem. We are given two vectors which are said to be perpendicular to each other. We can check the type of multiplication which can help us find the unknown ‘a’.
The vector product is given by –
$\overrightarrow{A}\times \overrightarrow{B}=\left| A \right|\left| B \right|\sin \theta \widehat{n}$
The scalar product is given by –
$\overrightarrow{A}.\overrightarrow{B}=\left| A \right|\left| B \right|\cos \theta$
From the above two methods, we can understand that, in the scalar product there exists a cosine component, which turns out to be zero when the vectors are perpendicular. Therefore, we can find the unknown ‘a’ by substituting in the scalar product of the given vectors A and B,

& \overrightarrow{A}=2\widehat{i}+3\widehat{j}+8\widehat{k} \\\ & \overrightarrow{B}=4\widehat{i}-4\widehat{j}+a\widehat{k} \\\ \end{aligned}$$ Now, let us find the dot product of these two vectors as – $$\begin{aligned} & \overrightarrow{A}.\overrightarrow{B}=\left| A \right|\left| B \right|\cos \theta \\\ & \theta =9{{0}^{0}}\Rightarrow \cos \theta =0 \\\ & \overrightarrow{A}=2\widehat{i}+3\widehat{j}+8\widehat{k} \\\ & \overrightarrow{B}=4\widehat{i}-4\widehat{j}+a\widehat{k} \\\ \end{aligned}$$ $$\begin{aligned} &\Rightarrow\text{}\overrightarrow{A}.\overrightarrow{B}=(2\widehat{i}+3\widehat{j}+8\widehat{k}).(4\widehat{i}-4\widehat{j}+a\widehat{k})=0 \\\ & \Rightarrow \text{ }8-12+8a=0 \\\ & \Rightarrow \text{ }8a=4 \\\ & \therefore \text{ }a=\dfrac{1}{2} \\\ \end{aligned}$$ From the above simple substitution, we have found the value of the unknown coefficient ‘a’. The required value for $$a=\dfrac{1}{2}$$. **So, the correct answer is “Option B”.** **Note:** We used the scalar product because the given two vectors were perpendicular to each other. If in a given case, the two vectors were parallel, we could have used the vector product instead, as we know the sine components reduces to zero and the product is zero.