Question

Let the position vectors of the points P and Q be $4\overrightarrow i + \overrightarrow j + \lambda \overrightarrow k$ and $2\overrightarrow i - \overrightarrow j + \lambda \overrightarrow k$ respectively. Vector $\overrightarrow i - \overrightarrow j + 6\overrightarrow k$ is perpendicular to the plane containing the origin and the points P and Q. Then $\lambda$equals.
$A)\dfrac{{ - 1}}{2}$
$B)\dfrac{1}{2}$
$C)1$
$D) - 1$

Answer 1

First, we will define the parallel axis and perpendicular. We know Parallel lines are the same lines from a distance apart. Perpendicular lines are the lines that will intersect at the right angle ${90^0}$. Vectors are the values with both magnitudes and direction.

Complete step-by-step solution:
From the given that, position vectors of the given points P and Q be $4\overrightarrow i + \overrightarrow j + \lambda \overrightarrow k$and$2\overrightarrow i - \overrightarrow j + \lambda \overrightarrow k$.
Which means, $\overrightarrow P = 4\overrightarrow i + \overrightarrow j + \lambda \overrightarrow k$ and $\overrightarrow Q = 2\overrightarrow i - \overrightarrow j + \lambda \overrightarrow k$. The origin point is O.
Now let us make the given vectors perpendicular to the plane of the origin O and the two points P, Q.
Which is the $\overrightarrow {OP} \times \overrightarrow {OQ}$ (perpendicular with the origin point from the given points)
Since the origin points of the vector are $\overrightarrow i + \overrightarrow j + \overrightarrow k$, and the points are given us $\overrightarrow P = 4\overrightarrow i + \overrightarrow j + \lambda \overrightarrow k$and$\overrightarrow Q = 2\overrightarrow i - \overrightarrow j + \lambda \overrightarrow k$.
Make these values into $3 \times 3$the matrix to find the new vector and that vector will be parallel to the given $\overrightarrow i - \overrightarrow j + 6\overrightarrow k$.
Thus, for the $3 \times 3$matrix, $\overrightarrow {OP} \times \overrightarrow {OQ}$ = \left| {\begin{array}{*{20}{c}} {\overrightarrow i }&{\overrightarrow j }&{\overrightarrow k } \\\ 4&1&\lambda \\\ 2&{ - 1}&\lambda \end{array}} \right| (concerning the matrix elements)
Hence taking the determinant we get,$\overrightarrow {OP} \times \overrightarrow {OQ}$ = \left| {\begin{array}{*{20}{c}} {\overrightarrow i }&{\overrightarrow j }&{\overrightarrow k } \\\ 4&1&\lambda \\\ 2&{ - 1}&\lambda \end{array}} \right| $\Rightarrow \overrightarrow i (\lambda + \lambda ) - \overrightarrow j (4\lambda - 2\lambda ) + \overrightarrow k ( - 4\lambda - 2\lambda )$
$\Rightarrow 2\lambda \overrightarrow i - 2\lambda \overrightarrow j + - 6\lambda \overrightarrow k$
Hence, we get, for $\overrightarrow {OP} \times \overrightarrow {OQ}$ $\Rightarrow 2\lambda \overrightarrow i - 2\lambda \overrightarrow j + - 6\lambda \overrightarrow k$, which is the new vector.
Now making the new vector $2\lambda \overrightarrow i - 2\lambda \overrightarrow j + - 6\lambda \overrightarrow k$ parallel to the given Vector $\overrightarrow i - \overrightarrow j + 6\overrightarrow k$ is perpendicular to the plane containing the origin.
Thus, for the parallel axis compare the two vectors with the corresponding vectors like for vector $i$ we have, two and one.
Therefore, comparing into equation format we get, $\dfrac{1}{{2\lambda }} = \dfrac{{ - 1}}{{ - 2\lambda }} = \dfrac{6}{{ - 6}}$ (representing vectors I, j, and k)
Simplifying the equation we get, $\dfrac{1}{{2\lambda }} = \dfrac{{ - 1}}{{ - 2\lambda }} = \dfrac{6}{{ - 6}} \Rightarrow \dfrac{1}{{2\lambda }} = \dfrac{1}{{2\lambda }} = - 1$
Hence if we substitute the lambda value as $\dfrac{{ - 1}}{2}$then we get $\Rightarrow \dfrac{1}{{2(\dfrac{{ - 1}}{2})}} = \dfrac{1}{{2(\dfrac{{ - 1}}{2})}} = - 1$
And thus, $- 1 = - 1 = - 1$ which satisfies the $\dfrac{1}{{2\lambda }} = \dfrac{{ - 1}}{{ - 2\lambda }} = \dfrac{6}{{ - 6}}$ (as all the values are same)
Hence $\lambda = \dfrac{{ - 1}}{2}$, the option $A)\dfrac{{ - 1}}{2}$ is correct.

Note: If we substitute the values like in options $B)\dfrac{1}{2}$ $C)1$ $D) - 1$ we don’t get the values of the equations to be equal.
Example apply $\lambda = 1$ in $\dfrac{1}{{2\lambda }} = \dfrac{{ - 1}}{{ - 2\lambda }} = \dfrac{6}{{ - 6}}$ then we get, $\dfrac{1}{2} = \dfrac{1}{2} = - 1$ (all the values are not same).
Hence after finding the equation form of the vectors, we need to substitute the correct value of the lambda.
Scalar means magnitudes only, but in vector values, we have both magnitude and direction