Question

If a vector $2\mathop i\limits^ \wedge + 3\mathop j\limits^ \wedge + 8\mathop k\limits^ \wedge $ is perpendicular to the vector $4\mathop j\limits^ \wedge - 4\mathop i\limits^ \wedge + \alpha \mathop k\limits^ \wedge $ then the value $\alpha $ is
$\left( a \right){\text{ }}\dfrac{1}{2}$
$\left( b \right){\text{ }}\dfrac{{ - 1}}{2}$
$\left( c \right){\text{ 1}}$
$\left( d \right){\text{ - 1}}$

Answer 1

Hint:
For solving this question we need to know one of the applications of dot product which is when the two vectors are perpendicular to each other, then their dot product is always zero. So by using this we will be able to solve this question.
Formula used:
If $\vec a$ is perpendicular to the $\vec b$ then,
$\vec a \cdot \vec b = 0$
Here, $\vec a\& \vec b$ , are the two vectors

Complete step by step solution:
For solving this question, we will let these two vectors named as
$ \Rightarrow \vec a = 2\mathop i\limits^ \wedge + 3\mathop j\limits^ \wedge + 8\mathop k\limits^ \wedge $
And, $\vec b = 4\mathop j\limits^ \wedge - 4\mathop i\limits^ \wedge + \alpha \mathop k\limits^ \wedge $
Now we will solve it, as we already know that their dot product will be zero, so writing the equation mathematically we get
$ \Rightarrow \left( {2\mathop i\limits^ \wedge + 3\mathop j\limits^ \wedge + 8\mathop k\limits^ \wedge } \right) \cdot \left( {4\mathop j\limits^ \wedge - 4\mathop i\limits^ \wedge + \alpha \mathop k\limits^ \wedge } \right) = 0$
Now on rearranging the equation in the standard form, we get
$ \Rightarrow \left( {2\mathop i\limits^ \wedge + 3\mathop j\limits^ \wedge + 8\mathop k\limits^ \wedge } \right) \cdot \left( { - 4\mathop i\limits^ \wedge + 4\mathop j\limits^ \wedge + \alpha \mathop k\limits^ \wedge } \right) = 0$
Now on solving the above equation, we get
\[ \Rightarrow - 8\mathop i\limits^ \wedge \cdot \mathop i\limits^ \wedge + 8\mathop i\limits^ \wedge \cdot \mathop j\limits^ \wedge + 2\alpha \mathop i\limits^ \wedge \cdot \mathop k\limits^ \wedge - 12\mathop j\limits^ \wedge \cdot \mathop i\limits^ \wedge + 12\mathop j\limits^ \wedge \cdot \mathop j\limits^ \wedge + 3\alpha \mathop j\limits^ \wedge \cdot \mathop k\limits^ \wedge - 32\mathop k\limits^ \wedge \cdot \mathop i\limits^ \wedge + 32\mathop k\limits^ \wedge \cdot \mathop j\limits^ \wedge + 8\alpha \mathop k\limits^ \wedge \cdot \mathop k\limits^ \wedge = 0\]
Now on solving the above equation using the concept of dot products, we get
$ \Rightarrow - 8 + 12 + 8\alpha = 0$
Now taking the constant term to the other side we get
$ \Rightarrow 8\alpha = - 4$
Now on further solving for the value, we get
$ \Rightarrow \alpha = \dfrac{{ - 1}}{2}$
Therefore, the value of $\alpha $ is $\dfrac{{ - 1}}{2}$.

Hence, the option $\left( b \right)$ is correct.

Note:
So for solving this we should know the dot product and cross-product multiplication, $\mathop i\limits^ \wedge \cdot \mathop i\limits^ \wedge = \mathop k\limits^ \wedge \cdot \mathop k\limits^ \wedge = \mathop j\limits^ \wedge \cdot \mathop j\limits^ \wedge = 1$ and similarly the $\mathop i\limits^ \wedge \cdot \mathop j\limits^ \wedge = \mathop j\limits^ \wedge \cdot \mathop k\limits^ \wedge = \mathop k\limits^ \wedge \cdot \mathop i\limits^ \wedge = 0$ . By using this we can solve the problem easily. Therefore, in the concept of vectors, if the vectors are similar then their product is always equal to one and if it is non-similar then their dot product will be zero. Similarly for the cross product of any vector with itself will always equal zero and with the other, the cross product will in two terms: when in cyclic pair \[\mathop i\limits^ \wedge \times \mathop j\limits^ \wedge = \mathop k\limits^ \wedge ,\mathop j\limits^ \wedge \times \mathop k\limits^ \wedge = \mathop i\limits^ \wedge ,\mathop k\limits^ \wedge \times \mathop i\limits^ \wedge = \mathop j\limits^ \wedge \] and if it’s in an anti-cyclic pair then it will be negative.