Question

Let $\overrightarrow a = \widehat i + \widehat j + \widehat k$ , $\overrightarrow c = \widehat j - \widehat k$ and a vector $\overrightarrow b $ be such that $\overrightarrow a \times \overrightarrow b = \overrightarrow c $ and $\overrightarrow a \cdot \overrightarrow b = 3$ . Then $|\overrightarrow b |$ equals:
A) $\sqrt {\dfrac{{11}}{3}} $
B) $\dfrac{{\sqrt {11} }}{3}$
C) $\dfrac{{11}}{{\sqrt 3 }}$
D) $\dfrac{{11}}{3}$

Answer 1

Hint:Assume the general expression for the vector $\overrightarrow b $ . The cross product of two vectors is a vector. Using this fact compare the vectors. Use the formula for modulus of a vector to find the final answer.

Complete step-by-step answer:
The data given in the problem is,
$\overrightarrow a \cdot \overrightarrow b = 3$.
$\overrightarrow a = \widehat i + \widehat j + \widehat k$
$\overrightarrow c = \widehat j - \widehat k$
Assume that the vector $\overrightarrow b $ is given by $\overrightarrow b = x\widehat i + y\widehat j + z\widehat k$.
It is given that the dot product $\overrightarrow a \cdot \overrightarrow b = 3$.
By using the formula for dot product, we can write the following:
$(\widehat i + \widehat j + \widehat k) \cdot (x\widehat i + y\widehat j + c\widehat k) = 3$
Therefore, simplifying the dot product we get,
$x + y + z = 3$
Now it is given that $\overrightarrow a \times \overrightarrow b = \overrightarrow c $ .
Therefore, using the formula for cross product we write:
$\left| {\begin{array}{*{20}{c}}
  {\widehat i}&{\widehat j}&{\widehat k} \\
  1&1&1 \\
  x&y&z
\end{array}} \right| = \widehat j - \widehat k$
Simplify the determinant,
$\widehat i\left( {z - y} \right) - \widehat j\left( {z - x} \right) + \widehat k\left( {y - x} \right) = \widehat j - \widehat k$
Comparing Left-hand side and right-hand side.
$z - y = 0$
This implies that $z = y$ .
Similarly,
$ - \left( {z - x} \right) = 1$
Implies that $x - z = 1$ .
And
$y - x = - 1$
Since we know that $z = y$ and $x + y + z = 3$ therefore, $x + 2y = 3$.
Now $x + 2y = 3$ and $y - x = - 1$.
Solving these two equations simultaneously we get, $y = \dfrac{2}{3}$ and $x = \dfrac{5}{3}$.
Now substituting these values in$x + y + z = 3$ we get the following:
$\dfrac{5}{3} + \dfrac{2}{3} + z = 3$
Simplifying for z we get the following:
$z = 3 - \dfrac{7}{3}$
Therefore, $z = \dfrac{2}{3}$ .
Therefore, the vector $\overrightarrow b $ is $\overrightarrow b = \dfrac{5}{3}\widehat i + \dfrac{2}{3}\widehat j + \dfrac{2}{3}\widehat k$ .
Now we will take modulus of the above vector as follow:
$\left| {\overrightarrow b } \right| = \sqrt {{{\left( {\dfrac{5}{3}} \right)}^2} + {{\left( {\dfrac{2}{3}} \right)}^2} + {{\left( {\dfrac{2}{3}} \right)}^2}} $
Simplify the squares and then we get:
$\left| {\overrightarrow b } \right| = \sqrt {\dfrac{{33}}{9}} $
We can simplify the square root by dividing numerator and denominator by 3.
Therefore, the value of the modulus is $\left| {\overrightarrow b } \right| = \sqrt {\dfrac{{11}}{3}} $.

So, the correct answer is “Option A”.

Note:Here the important point to note is that the cross product is always vector. The modulus of the vector is always a scalar. We calculate the cross product and use it to find the component of the vector.