Question

Vector $\overrightarrow c$ is perpendicular to vectors $\overrightarrow a = (2, - 3,1)$ and $\overrightarrow b = (1, - 2,3)$ satisfies the condition$\overrightarrow c .(\overrightarrow i + 2\overrightarrow j - 7\overrightarrow k ) = 10$. Then vector $\overrightarrow c$is equal to
A. $(7,5,1)$
B. $( - 7, - 5, - 1)$
C. $(1,1, - 1)$
D. $(-1, -1, - 1)$

Answer 1

Two vectors A and B are perpendicular if and only if their scalar product is equal to zero.
That is $(\overrightarrow A .\overrightarrow B ) = 0$
Using this condition of perpendicular vectors we can find the value of$\overrightarrow c$and we will verify the answer found using the given second condition.

Complete step-by-step answer:
It is given that the vector $\overrightarrow c$ is perpendicular to vectors $\overrightarrow a = (2, - 3,1)$ and $\overrightarrow b = (1, - 2,3)$.
It is also given that$\overrightarrow c .(\overrightarrow i + 2\overrightarrow j - 7\overrightarrow k ) = 10$.
Since, $\overrightarrow c$ is perpendicular to $\overrightarrow a$ and$\overrightarrow b$.
We can say that$\overrightarrow c$is the constant multiple of the cross product of $\overrightarrow a$ and$\overrightarrow b$,
That is $\overrightarrow c = \lambda (\overrightarrow a \times \overrightarrow b )$… (1)
Where, $\lambda$ is known as the constant multiplied with the cross product.
Now let us find the cross product of $\overrightarrow a$ and$\overrightarrow b$,
(\overrightarrow a \times \overrightarrow b ) = \left( {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\\2&{ - 3}&1\\\1&{ - 2}&3\end{array}} \right)
Let us solve the above matrix to find the value of cross product, we get,
$(\overrightarrow a \times \overrightarrow b ) = ( - 9 + 2)\widehat i - (6 - 1\widehat {)j} + ( - 4 + 3)\widehat k$
Which implies,
$(\overrightarrow a \times \overrightarrow b ) = - 7\widehat i - 5\widehat j - \widehat k$
Let us substitute the value of cross product in (1), we get,
$\overrightarrow c = \lambda ( - 7\widehat i - 5\widehat j - \widehat k)$
We have a second condition that,
$\overrightarrow c .(\overrightarrow i + 2\overrightarrow j - 7\overrightarrow k ) = 10$
Now let us substitute the value of$\overrightarrow c$in the above condition, we get,
$\lambda ( - 7\widehat i - 5\widehat j - \widehat k).(\overrightarrow i + 2\overrightarrow j - 7\overrightarrow k ) = 10$
By the definition of scalar product we multiply the like vectors and by simplifying we get,
$\lambda ( - 7 - 10 + 7) = 10$
That is on solving we get,
$- 10\lambda = 10$
Let us divide by $- 10$ on both sides we get,
$\lambda = - 1$
Let us substitute the above value in the value of $\overrightarrow c$ we get,
$\overrightarrow c = (7\widehat i + 5\widehat j + \widehat k)$

So, the correct answer is “Option A”.

Additional information: The cross product $(\overrightarrow a \times \overrightarrow b )$ is defined as a vector c that is perpendicular (orthogonal) to both “a” and “b”, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.
The cross product is defined by the formula
(\overrightarrow a \times \overrightarrow b ) = \left( {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\\\{{a_1}}&{{a_2}}&{{a_3}}\\\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right)

Note: Here in the given options, option (c) also satisfies the second condition $\overrightarrow c .(\overrightarrow i + 2\overrightarrow j - 7\overrightarrow k ) = 10$ but the option (c) fails with the first condition that it is not perpendicular to $\overrightarrow a$ and $\overrightarrow b$.