Question

Let $O$ be the origin. Let $OP = xi + yj - k$ and $OQ = - i + 2j + 3xk$ , $x,y \in R$ , $x \succ 0$ , be such that \[\left| {PQ} \right| = \sqrt {20} \] and the vector $OP$ is perpendicular to $OQ$ . if $OR = 3i + zj - 7k \in R$ , is coplanar with $OP$ and $OQ$ , then the value of ${x^2} + {y^2} + {z^2}$ is
A) $2$
B) $9$
C) $1$
D) $7$

Answer 1

Hint:Since it is given that the vector $OP$ is perpendicular to $OQ$, so their dot product will be zero. Using this we can find a relation between $x$ and $y$. Here, we know the magnitude of $PQ$ so we can equate it with the magnitude formula of vector $PQ$ using vectors $OP$ and $OQ$. This way we get the value of $x$ and find the value of $y$ from the relation we got between $x$ and $y$. Also, it is given that $OR$ is coplanar with $OP$ and $OQ$ so when we keep these vectors in the matrix form, their determinant comes out to be 0 and we get the value of $z$. Hence, get the value of the required expression by substituting the values of $x,y$, and $z$.

Formula Used:
 ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ and ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$

Complete step by step Solution:
 Since, $\overrightarrow {OP} $ is perpendicular to $\overrightarrow {OQ} $ therefore, their dot product should be zero.
 $\overrightarrow {OP} .\overrightarrow {OQ} = 0$
Substituting their values from the question, we get
 $(xi + yj - k).( - i + 2j + 3xk) = 0$
 $ - x + 2y - 3x = 0$
This gives us $2y - 4x = 0$
Thus, $y = 2x$ ...(1)
Now we know that vector $PQ$ can be found as: $\overrightarrow {PQ} = \overrightarrow {OQ} - \overrightarrow {OP} $
This implies that $\overrightarrow {PQ} = - i + 2j + 3xk - \left( {xi + yj - k} \right)$
  $\overrightarrow {PQ} = \left( { - x - 1} \right)i + \left( {2 - y} \right)j + (3x + 1)k$
Now it is given that \[\left| {PQ} \right| = \sqrt {20} \]
And also \[\left| {PQ} \right| = \sqrt {{{\left( { - x - 1} \right)}^2} + {{\left( {2 - y} \right)}^2} + {{\left( {3x + 1} \right)}^2}} \]
This gives us \[\sqrt {{{\left( { - x - 1} \right)}^2} + {{\left( {2 - y} \right)}^2} + {{\left( {3x + 1} \right)}^2}} = \sqrt {20} \] ...(2)
Squaring both sides and substituting equation (1) in equation (2), we get
 \[{\left( { - x - 1} \right)^2} + {\left( {2 - 2x} \right)^2} + {\left( {3x + 1} \right)^2} = 20\]
Taking the negative sign out of square in the first term and 2 in the second term, we get
 \[{\left( {x + 1} \right)^2} + 4{\left( {1 - x} \right)^2} + {\left( {3x + 1} \right)^2} = 20\]
Using identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ and ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ , we get
 ${x^2} + 1 + 2x + 4 + 4{x^2} - 8x + 9{x^2} + 1 + 6x = 20$
The $x$ terms cancel out and by solving the rest of the terms, we get
 $14{x^2} = 20 - 6$
 $14{x^2} = 14$
We get ${x^2} = 1$
This implies that $x = 1, - 1$ but since it is given in the question that $x \succ 0$ therefore $x$ cannot be equal to -1.
Therefore, $x = 1$ ...(3)
Substituting value of $x$ in equation (2), we get $y = 2$ . ...(4)
Now, we are given that $OR$ is coplanar with $OP$ and $OQ$ so when we keep these vectors in the matrix form, their determinant comes out to be 0.
 \[\left| {\left. {\begin{array}{*{20}{c}}
  x&y&{ - 1} \\
  { - 1}&2&{3x} \\
  3&z&{ - 7}
\end{array}} \right|} \right. = 0\]
Substitute equations (3) and (4) in the determinant, and we get
 \[\left| {\left. {\begin{array}{*{20}{c}}
  1&2&{ - 1} \\
  { - 1}&2&3 \\
  3&z&{ - 7}
\end{array}} \right|} \right. = 0\]
Solving this we get,
 \[1\left[ {\left( {2 \times - 7} \right) - 3z} \right] - 2\left[ {\left( { - 1 \times - 7} \right) - \left( {3 \times 3} \right)} \right] + \left( { - 1} \right)\left[ {\left( { - 1 \times z} \right) - \left( {2 \times 3} \right)} \right] = 0\]
 $\left( { - 14 - 3z} \right) - 2\left( {7 - 9} \right) - \left( { - z - 6} \right) = 0$
Open the brackets and solve,
 $ - 14 - 3z - 14 + 18 + z + 6 = 0$
 $ - 2z - 4 = 0$
Thus, we get $z = - 2$
Hence, we can find the value of the required equation.
 ${x^2} + {y^2} + {z^2} = {1^2} + {2^2} + {\left( { - 2} \right)^2}$
 ${x^2} + {y^2} + {z^2} = 9$

Therefore, the correct option is B.

Note: More such questions can be made by either changing the way of framing the question or by changing the conditions like the two vectors are perpendicular or coplanar or start from the same point etc. These questions can also include trigonometric terms and concepts.