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

Hint: 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 \[\overrightarrow b \].