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Mathematics Question on Vector Algebra

Let, a⃗=−i^−k^,b⃗=−i^+j^\vec {a} = - \hat {i} - \hat {k}, \vec {b} = - \hat {i} +\hat {j} and c⃗=i^+2j^+3k^\vec {c} = \hat {i}+2\hat {j}+3\hat {k} be three given vectors. If r⃗\vec {r} is a vector such that r⃗×b⃗=c⃗×b⃗\vec {r} \times \vec {b} = \vec {c} \times \vec{b} and r⃗⋅a⃗=0\vec {r} \cdot \vec{a} = 0 , then the value of r⃗⋅b⃗\vec {r} \cdot \vec {b} is

A

3

B

6

C

9

D

12

Answer

9

Explanation

Solution

Let r=x1i^+x2j^+x3k^r=x_{1}\hat{i}+x_{2} \hat{j} +x_{3} \hat{k}
Now, c×b=∣i^j^k^ 123 −110∣c\times b=\left|\begin{matrix}\hat{i}&\hat{j}&\hat{k}\\\ 1&2&3\\\ -1&1&0\end{matrix}\right|
=i^(0−3)−j^(0+3)+k^(1+2)=\hat{i} \left(0-3\right)-\hat{j} \left(0+3\right)+\hat{k}\left(1+2\right)
=−3i^−3j^+3k^=-3 \hat{i}-3\hat{j}+3\hat{k}
and r×b=∣i^j^k^ x1x2x3 −110∣r\times b=\left|\begin{matrix}\hat{i}&\hat{j}&\hat{k}\\\ x_{1}&x_{2}&x_{3}\\\ -1&1&0\end{matrix}\right|
=i^(0−x3)−j^(0+x3)+k^(x1+x2)=\hat{i}\left(0-x_{3}\right)-\hat{j} \left(0+x_{3}\right)+\hat{k} \left(x_{1}+x_{2}\right)
=−x3 i^−x3 j^(x1+x2)k^=-x_{3}\, \hat{i} -x_{3}\, \hat{j} \left(x_{1}+x_{2}\right) \hat{k}
Also, r⋅a=0r\cdot a=0
⇒(x1i^+x2j^+x3k^)⋅(−i^−k^)=0\Rightarrow \left(x_{1} \hat{i}+x_{2} \hat{j} +x_{3}\hat{k}\right)\cdot\left(-\hat{i}-\hat{k}\right)=0
⇒−x1−x3=0…(i)\Rightarrow -x_{1}-x_{3}=0 \ldots\left(i\right)
But r×b=c×br\times b=c\times b
∴−x3i^−x3j^+k^(x1+x2)=−3i^−3j^+3k^\therefore -x_{3} \hat{i}-x_{3} \hat{j} +\hat{k} \left(x_{1}+x_{2}\right)=-3 \hat{i} -3 \hat{j} +3\hat{k}
On comparing both sides, we get
x3=3x_{3}=3 and x1+x2=3…(ii)x_{1}+x_{2}=3 \ldots\left(ii\right)
On solving Eqs. (i)\left(i\right) and (ii)\left(ii\right), we get
x1=−x3=−3x_{1}=-x_{3}=-3 and x2=6x_{2}=6
Now, r⋅b=(x1i^+x2j^+x3k^)⋅(−i^+j^)r\cdot b=\left(x_{1}\hat{i} +x_{2} \hat{j} +x_{3}\hat{k}\right) \cdot\left(-\hat{i}+\hat{j}\right)
=−x1+x2=−(−3)+6=-x_{1}+x_{2}=-\left(-3\right)+6
=9=9

So, the correct answer is (C): 9