Question

In a triangle ABC, the sides AB and AC are represented by the vectors $3\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k}$ respectively. Calculate the angle $\angle ABC$.
A. ${{\cos }^{-1}}\sqrt{\dfrac{5}{11}}$
B. ${{\cos }^{-1}}\sqrt{\dfrac{6}{11}}$
C. ${{90}^{\circ }}-{{\cos }^{-1}}\sqrt{\dfrac{5}{11}}$
D. ${{180}^{\circ }}-{{\cos }^{-1}}\sqrt{\dfrac{5}{11}}$

Answer 1

Position vector is a straight line having one end fixed to a body and the other end attached to a moving point and it is used to describe the position of the point relative to the body. As the particle moves, the position vector corresponding to that particle will also change in length or in direction or in both length and direction. To solve this question, we have to keep the following formulas in mind
Magnitude of a vector quantity is given by $\left| \overset{\to }{\mathop{PQ}}\, \right|=\sqrt{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}$
The scalar product or a dot product between any two vectors is given by $\overset{\to }{\mathop{CB}}\,.\overset{\to }{\mathop{AB}}\,=\left| \overset{\to }{\mathop{AB}}\, \right|\left| \overset{\to }{\mathop{CB}}\, \right|\cos \theta$ where $\theta$ .is the angle between the two vectors.

Complete step-by-step answer:
If $\overset{\to }{\mathop{PQ}}\,$ is a given vector where $\overset{\to }{\mathop{PQ}}\,=\left( x\hat{i}+y\hat{j}+z\hat{k} \right)$ then $\left| \overset{\to }{\mathop{PQ}}\, \right|$ is given by,
$\left| \overset{\to }{\mathop{PQ}}\, \right|=\sqrt{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}$
Now it is given that,
$\overset{\to }{\mathop{AB}}\,=3\hat{i}+\hat{j}+\hat{k}$
$\overset{\to }{\mathop{AC=}}\,\hat{i}+2\hat{j}+\hat{k}$
The position vector from B to C will be given by
$\overset{\to }{\mathop{CB}}\,=\overset{\to }{\mathop{AB}}\,-\overset{\to }{\mathop{AC}}\,$
$=2\hat{i}-\hat{j}$
Then $\angle ABC$ is angle between AB and CB will be given by
$\overset{\to }{\mathop{CB}}\,.\overset{\to }{\mathop{AB}}\,=\left| \overset{\to }{\mathop{AB}}\, \right|\left| \overset{\to }{\mathop{CB}}\, \right|\cos \theta$
$\left( 3\hat{i}+\hat{j}+\hat{k} \right).\left( \hat{i}+2\hat{j}+\hat{k} \right)=\left| \sqrt{5} \right|\left| \sqrt{11} \right|\cos \theta$
$5=\left| \sqrt{5} \right|\left| \sqrt{11} \right|\cos \theta$
$\dfrac{\left| \sqrt{5} \right|}{\left| \sqrt{11} \right|}=\cos \theta$
$\theta ={{\cos }^{-1}}\sqrt{\dfrac{5}{11}}$.
Hence, the correct answer is ${{\cos }^{-1}}\sqrt{\dfrac{5}{11}}$ which is option A.

Note: Vector quantity in physics is defined as a quantity that has both magnitude as well as direction. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the magnitude of the quantity. Some examples of vector quantities are force, displacement, electric potential etc. Position vector is a vector quantity representing the position of a particle with its direction in a Cartesian plane.