Question

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

Answer 1

At first, we rewrite the equation as $a\times \left( 2i+3j+4k \right)=-\left( b\times \left( 2i+3j+4k \right) \right)$ since $y\times x$ can be written as $-\left( x\times y \right)$ . Then, we write it as $\left( a+b \right)\times \left( 2i+3j+4k \right)=0$ . We can then tell that $a+b$ and $2i+3j+4k$ are parallel to each other. We then find $\left| 2i+3j+4k \right|$ which comes out as $\sqrt{29}$ , same as that of $\left| a+b \right|$ . So, $a+b$ and $2i+3j+4k$ are same. So, $\left( a+b \right).\left( -7i+2j+3k \right)$ is nothing but $\left( 2i+3j+4k \right).\left( -7i+2j+3k \right)$ . Solving this gives the answer.

Complete step by step answer:
The equation of a and b that we are provided with in this problem is,
$a\times \left( 2i+3j+4k \right)=\left( 2i+3j+4k \right)\times b$
Now, we know that the vector $x\times y$ is perpendicular to both the vectors x and y. Also, we know that the vector $y\times x$ is perpendicular to both the vectors x and y. The only difference between the vectors $x\times y$ and $y\times x$ is their direction. They are opposite to each other. That means that $y\times x$ can be written as $-\left( x\times y \right)$ . In a similar way, we can write $\left( 2i+3j+4k \right)\times b$ as $-\left( b\times \left( 2i+3j+4k \right) \right)$ . The equation becomes,
$\begin{aligned} & \Rightarrow a\times \left( 2i+3j+4k \right)=-\left( b\times \left( 2i+3j+4k \right) \right) \\\ & \Rightarrow a\times \left( 2i+3j+4k \right)+b\times \left( 2i+3j+4k \right)=0 \\\ \end{aligned}$
Now, we know that the vector $\left( x+y \right)\times z$ can be written as $\left( x\times z \right)+\left( y\times z \right)$ . So, the equation becomes,
$\Rightarrow \left( a+b \right)\times \left( 2i+3j+4k \right)=0$
Now, we know that the magnitude of the vector $x\times y$ is $\left| x \right|\left| y \right|\sin \theta$ where, $\theta$ is the angle between the two vectors x and y. So, we can write,
$\begin{aligned} & \Rightarrow \left| \left( a+b \right)\times \left( 2i+3j+4k \right) \right|=\left| 0 \right| \\\ & \Rightarrow \left| a+b \right|\left| 2i+3j+4k \right|\sin \theta =0 \\\ \end{aligned}$
Now, we are given that $\left| a+b \right|=\sqrt{29}$ and $\left| 2i+3j+4k \right|$ is also not clearly zero. This means that $\sin \theta$ must be zero, or $\theta$ must be zero. This means that the two vectors $a+b$ and $2i+3j+4k$ are parallel to each other. Let us calculate $\left| 2i+3j+4k \right|$ .
$\Rightarrow \left| 2i+3j+4k \right|=\sqrt{{{2}^{2}}+{{3}^{2}}+{{4}^{2}}}=\sqrt{29}$
This means that $a+b$ and $2i+3j+4k$ are one and the same as both their magnitudes and directions are the same. We know that $\left( {{x}_{1}}i+{{x}_{2}}j+{{x}_{3}}k \right).\left( {{y}_{1}}i+{{y}_{2}}j+{{y}_{3}}k \right)={{x}_{1}}{{y}_{1}}+{{x}_{2}}{{y}_{2}}+{{x}_{3}}{{y}_{3}}$ . So,
$\begin{aligned} & \Rightarrow \left( a+b \right).\left( -7i+2j+3k \right)=\left( 2i+3j+4k \right).\left( -7i+2j+3k \right) \\\ & \Rightarrow \left( a+b \right).\left( -7i+2j+3k \right)=2\left( -7 \right)+3\left( 2 \right)+4\left( 3 \right) \\\ & \Rightarrow \left( a+b \right).\left( -7i+2j+3k \right)=4 \\\ \end{aligned}$
Thus, we can conclude that $\left( a+b \right).\left( -7i+2j+3k \right)$ is $4$ .

So, the correct answer is “Option C”.

Note: We must be very clear about the basic concepts of vectors. Knowing only the various formulae will not help in the long run. Vectors involve a lot of intuition, which develops only when we solve a lot of problems.