Question

If $\overrightarrow a$ and $\overrightarrow b$ are vectors such that $\left| {\overrightarrow {a{\text{ }}} + {\text{ }}\overrightarrow b } \right| = \sqrt {29}$ and$\overrightarrow a {\text{ }} \times \left( {2\widehat i + 3\widehat j + 4\widehat k} \right){\text{ = }}\left( {2\widehat i + 3\widehat j + 4\widehat k} \right){\text{ }} \times {\text{ }}\overrightarrow b$, then a possible value of $\left( {\overrightarrow a + \overrightarrow b } \right).\left( {-7\widehat i + 2\widehat j{\text{ + }}3\widehat k} \right)$ is
A) 0
B) 3
C) ±4
D) 8

Answer 1

There are two equalities given in the question and you are asked to find third. We have to try to use the first two so that we can derive the third. Here I used the second equality and substituted the value of first in the second equation. With the resultant equation, I tried to solve the third equation.

Complete step-by-step answer:
Given, $\overrightarrow a {\text{ }} \times \left( {2\widehat i + 3\widehat j + 4\widehat k} \right){\text{ = }}\left( {2\widehat i + 3\widehat j + 4\widehat k} \right){\text{ }} \times {\text{ }}\overrightarrow b$
The two special properties of cross product of vector are $\left( {b \times a = - a \times b,{\text{ }}and{\text{ }}a \times a = 0} \right)$, we'll use the first one here:-
$\overrightarrow a {\text{ }} \times \left( {2\widehat i + 3\widehat j + 4\widehat k} \right){\text{ = - }}\overrightarrow b \times \left( {2\widehat i + 3\widehat j + 4\widehat k} \right){\text{ }}$
$\Rightarrow \left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {2\widehat i + 3\widehat j + 4\widehat k} \right){\text{ = 0}}$
The above statement proves that $\left( {\overrightarrow a + \overrightarrow b } \right)$ and $\left( {2\widehat i + 3\widehat j + 4\widehat k} \right)$are two parallel vectors as their cross product is zero.
$\Rightarrow \left( {\overrightarrow a + \overrightarrow b } \right) = \lambda \left( {2\widehat i + 3\widehat j + 4\widehat k} \right)$
$\Rightarrow \left| {\left( {\overrightarrow a + \overrightarrow b } \right)} \right| = \left| {\lambda \left( {2\widehat i + 3\widehat j + 4\widehat k} \right)} \right|$
$\left| {\left( {2\widehat i + 3\widehat j + 4\widehat k} \right)} \right| = \sqrt {{2^2} + {3^2} + {4^2}} = \sqrt {29}$
And, the value of $\left| {\overrightarrow {a{\text{ }}} + {\text{ }}\overrightarrow b } \right| = \sqrt {29}$ already given in the question.
By putting both these value in the above question, we get
$\Rightarrow \sqrt {29} = {\lambda ^2}\sqrt {29}$
$\Rightarrow \lambda = \pm 1$
Thus we get, $\left| {\overrightarrow {a{\text{ }}} + {\text{ }}\overrightarrow b } \right| = \pm \left| {\left( {2\widehat i + 3\widehat j + 4\widehat k} \right)} \right|$
Now we have calculate the value of :
(Since we know the dot product of unit vectors, we can simplify the dot product formula to
a⋅b=a1b1+a2b2+a3b3)
$\left( {\overrightarrow a + \overrightarrow b } \right).\left( {-7\widehat i + 2\widehat j{\text{ + }}3\widehat k} \right)$

$$= \left( {2\widehat i + 3\widehat j + 4\widehat k} \right).\left( {-7\widehat i + 2\widehat j{\text{ + }}3\widehat k} \right) \\\ = \pm \left( { - 14 + 6 + 12} \right) = \pm 4 \\\$$

So, option (C) is the correct answer.

Note: We must only multiply a vector with other scalar and vector quantities. The vector multiplication, however, is not an unique mathematical construct like scalar multiplication. The multiplication depends on the nature of quantities (vector or scalar) and on the physical process, necessitating scalar or vector multiplication. Vector dot product and cross product are two types of vector product, the basic difference between dot product and the cross product is that in dot product, the product of two vectors is equal to scalar quantity while in the cross product, the product of two vectors is equal to vector quantity.