Question

${a_1}\hat i + {a_2}\hat j$ is a unit vector perpendicular to $4\hat i - 3\hat j$ if
\left( A \right){a_1} = 0.6,{a_2} = 0.8 \\\ \left(B \right){a_1} = 3,{a_2} = 4 \\\ \left( C \right){a_1} = 0.8,{a_2} = 0.6 \\\ \left( D \right){a_1} = 4,{a_2} = 3 \\\
If $\vec a = 2\hat i - 3\hat j$ and $\vec b = 2\hat j + 3\hat k$ , then $\left( {\vec a + \vec b} \right) \cdot \left( {\vec a - \vec b} \right) =$
\left( A \right)0 \\\ \left( B \right) - 8 \\\ \left( C \right)9 \\\ \left( D \right)1 \\\

Answer 1

Hint : In order to solve the first part, we are going to form two equations, one for the magnitude of the unit vector ${a_1}\hat i + {a_2}\hat j$ and one for the dot product of the vectors ${a_1}\hat i + {a_2}\hat j$ and $4\hat i - 3\hat j$ , to solve for the values of ${a_1}$ and ${a_2}$ . In the second part, we are going to first find the vector sum $\left( {\vec a + \vec b} \right)$ and the vector difference $\left( {\vec a - \vec b} \right)$ , and then their dot product.
Formula used: The magnitude of a unit vector ${a_1}\hat i + {a_2}\hat j$ is given by
$\sqrt {{a_1}^2 + {a_2}^2} = 1$
The sum of the two vectors $x\hat i + y\hat j + z\hat k$ and $a\hat i + b\hat j + c\hat k$ is
$\left( {a + x} \right)\hat i + \left( {b + y} \right)\hat j + \left( {z + c} \right)\hat k$
The difference of the two vectors $x\hat i + y\hat j + z\hat k$ and $a\hat i + b\hat j + c\hat k$ is
$\left( {a - x} \right)\hat i + \left( {b - y} \right)\hat j + \left( {z - c} \right)\hat k$
The dot product of the two vectors $x\hat i + y\hat j + z\hat k$ and $a\hat i + b\hat j + c\hat k$ is
$ax + by + cz$

Complete Step By Step Answer:
It is given that ${a_1}\hat i + {a_2}\hat j$ is a unit vector
This implies that the magnitude of the vector is equal to $1$
Mathematically, we can write
$\sqrt {{a_1}^2 + {a_2}^2} = 1 - - - \left( 1 \right)$
Now, it is also given that ${a_1}\hat i + {a_2}\hat j$ is a unit vector perpendicular to $4\hat i - 3\hat j$
This implies
$\left( {{a_1}\hat i + {a_2}\hat j} \right) \cdot \left( {4\hat i - 3\hat j} \right) = 0$
Hence the equation becomes
4{a_1} - 3{a_2} = 0 \\\ \Rightarrow {a_1} = \dfrac{3}{4}{a_2} \\\
Using this in the equation $\left( 1 \right)$
\Rightarrow {\left( {\dfrac{3}{4}{a_2}} \right)^2} + {a_2}^2 = 1 \\\ \Rightarrow \dfrac{9}{{16}}{a_2}^2 + {a_2}^2 = 1 \\\ \Rightarrow \dfrac{{25}}{{16}}{a_2}^2 = 1 \\\ \Rightarrow {a_2}^2 = \dfrac{{16}}{{25}} \\\ \Rightarrow {a_2} = \dfrac{4}{5} = 0.8 \\\
Now using this value to get the value for ${a_1}$
$\Rightarrow {a_1} = \dfrac{3}{4} \times 0.8 = 0.6$
Thus, the option $\left( A \right){a_1} = 0.6,{a_2} = 0.8$ is correct.
In part $2$ , it is given that $\vec a = 2\hat i - 3\hat j$ and $\vec b = 2\hat j + 3\hat k$
Now,
\vec a + \vec b = \left( {2\hat i - 3\hat j} \right) + \left( {2\hat j + 3\hat k} \right) = 2\hat i - \hat j + 3\hat k \\\ \vec a - \vec b = \left( {2\hat i - 3\hat j} \right) - \left( {2\hat j + 3\hat k} \right) = 2\hat i - 5\hat j - 3\hat k \\\
Then, $\left( {\vec a + \vec b} \right) \cdot \left( {\vec a - \vec b} \right)$ can be computed as
$\left( {\vec a + \vec b} \right) \cdot \left( {\vec a - \vec b} \right) = \left( {2\hat i - \hat j + 3\hat k} \right) \cdot \left( {2\hat i - 5\hat j - 3\hat k} \right) = 4 + 5 - 9 = 0$
Hence, option $\left( A \right)0$ is the correct option.

Note :
It is important to note that as in the first part, we are given with the magnitude of the vector ${a_1}\hat i + {a_2}\hat j$ , many students miss that and are able to form just one equation. However, with two equations, it is very simple to solve this. In the second part, the sum and the difference vectors are calculated directly and the dot product is done.