Question

Let $\overrightarrow u $ be a vector coplanar with the vectors $\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k$ and $\overrightarrow b = \widehat j + \widehat k$ . If $\overrightarrow u $ is perpendicular to $\overrightarrow a $ and $\overrightarrow u .\overrightarrow b = 24$ , then ${\left| u \right|^2}$ is equal to
A. $256$
B. $84$
C. $336$
D. $315$

Answer 1

Hint: At first, we need to assume $\overrightarrow u $ as a variable vector. Then, get an equation in terms of that variables by using the statement that $\overrightarrow u $is a vector coplanar with the vectors $\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k$ and $\overrightarrow b = \widehat j + \widehat k$ . Form the second equation by using the information that$\overrightarrow u $ is perpendicular to $\overrightarrow a $ and the third equation by $\overrightarrow u .\overrightarrow b = 24$ and solve these equations.

Complete step-by-step solution:
Here in this problem, we are given three coplanar vectors $\overrightarrow u ,\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k{\text{ and }}\overrightarrow b = \widehat j + \widehat k$. We also know that $\overrightarrow u $ is perpendicular to $\overrightarrow a $ and $\overrightarrow u .\overrightarrow b = 24$. Using this given information, we need to find the value of ${\left| u \right|^2}$
Let us assume that $\overrightarrow u = x\widehat i + y\widehat j + z\widehat k$
We already know $\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k$ and $\overrightarrow b = \widehat j + \widehat k$
Now we know that it is coplanar with $\overrightarrow a ,\overrightarrow b $. It means that the determinant formed by $\overrightarrow u ,\overrightarrow a ,\overrightarrow b $is equal to zero.
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&y&z \\
  2&3&{ - 1} \\
  0&1&1
\end{array}} \right| = 0$
The determinant can be solved as $\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right| = a\left( {ei - hf} \right) - b\left( {di - fg} \right) + c\left( {dh - eg} \right)$
Therefore, we get:
$ \Rightarrow x(3 + 1) - y(2 + 0) + z(2 - 0) = 0 \Rightarrow 4x - 2y + 2z = 0$$ - - - - (1)$
Now form the second equation by using the information that $\overrightarrow u $ is perpendicular to $\overrightarrow a $
This statement implies that:
$ \Rightarrow (x\widehat i + y\widehat j + z\widehat k) \cdot (2\widehat i + 3\widehat j - \widehat k) = 0$ , which is the dot or scalar product of vector $\overrightarrow u $ and $\overrightarrow a $
Now let’s solve it to get:
$ \Rightarrow 2x + 3y - z = 0$$ - - - - (2)$
At last, we will use $\overrightarrow u .\overrightarrow b = 24$
$ \Rightarrow (x\widehat i + y\widehat j + z\widehat k) \cdot (\widehat j + \widehat k) = 24 \Rightarrow y + z = 24$$ - - - - - (3)$
Adding (2) and (3), we get
$ \Rightarrow 2x + 3y - z + y + z = 0 + 24 \Rightarrow 2x + 4y = 24$
On dividing the whole equation by two, we get:
$ \Rightarrow x + 2y = 12$$ - - - - - (4)$
Multiplying equation (3) by $2$and subtracting it to (2)
$ \Rightarrow 4x - 4y = 48 \Rightarrow x - y = - 12$$ - - - - - - (5)$
Subtracting equation (5) from (4), we get:
$ \Rightarrow 3y = 24 \Rightarrow y = 8$
Using it in equation (5), we obtain $x = - 4$
Substituting it in (2), we can find the value for ‘z’
$ \Rightarrow (2)( - 4) + (3)(8) - z = 0 \Rightarrow z = 16$
Therefore, we can write the vector $\overrightarrow u = - 4\widehat i + 8\widehat j + 16\widehat k$
The expression \[\left| u \right|\] represents the magnitude of the vector, which can be given by the square root of the sum of the squares of the direction ratios, i.e.
$ \Rightarrow \left| {\overrightarrow u } \right| = \sqrt {{{( - 4)}^2} + {8^2} + {{16}^2}} = \sqrt {16 + 64 + 256} = \sqrt {336} $
So the required expression ${\left| u \right|^2}$ can be written as:
$ \Rightarrow {\left| u \right|^2} = \left| u \right| \times \left| u \right| = {\left( {\sqrt {336} } \right)^2} = 336$
Thus, the value of the required expression is ${\left| u \right|^2} = 336$

Hence, the option (C) is the correct answer.

Note: The key to solving this question was the formation of the equation $\left( 1 \right)$ which is to figure out the meaning of vectors being coplanar. The dot product of two vectors $\vec a{\text{ and }}\vec b$ is given by $\vec a \cdot \vec b = \left| {\vec a} \right|\left| {\vec b} \right|\cos \theta $ , where ‘theta’ represents the angle between two vectors. So when we find the dot product of two perpendicular vectors, we get $\cos \dfrac{\pi }{2} = 0 \Rightarrow \vec a \cdot \vec b = \left| a \right|\left| b \right| \times 0 = 0$ .