Question

If A and B are the points (2, 1, -2), (3, -4, 5) then the angle OA makes with OB is
(a) ${{\cos }^{-1}}\left( \dfrac{4\sqrt{2}}{15} \right)$
(b) ${{\cos }^{-1}}\left( \dfrac{4}{15\sqrt{2}} \right)$
(c) ${{\cos }^{-1}}\left( \dfrac{8\sqrt{2}}{15} \right)$
(d) $\dfrac{\pi }{2}$

Answer 1

To solve this question we will use the formula of dot product of two vectors which is given as, $\overset{\to }{\mathop{P}}\,.\overset{\to }{\mathop{Q}}\,=\left| \overset{\to }{\mathop{P}}\, \right|\left| \overset{\to }{\mathop{Q}}\, \right|\cos \theta$, where $\theta$ is angle between vectors, $\overset{\to }{\mathop{P}}\,$ and $\overset{\to }{\mathop{Q}}\,$, $\left| \overset{\to }{\mathop{P}}\, \right|$ is length of vector $\overset{\to }{\mathop{P}}\,$ also $\left| \overset{\to }{\mathop{P}}\, \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$, where $\overset{\to }{\mathop{P}}\,=\left( a,b,c \right)$. Also the dot product of two vectors is given by $\left( a\overset{\wedge }{\mathop{i}}\,+b\overset{\wedge }{\mathop{j}}\,+c\overset{\wedge }{\mathop{k}}\, \right).\left( x\overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{j}}\,+z\overset{\wedge }{\mathop{k}}\, \right)=ax+by+cz$.

Complete step-by-step solution:
Given that, A = (2, 1, -2) and B = (3, -4, 5).
We have to find angles between $\overset{\to }{\mathop{OA}}\,$ and $\overset{\to }{\mathop{OB}}\,$ to do that first of all we will calculate $\overset{\to }{\mathop{OA}}\,$ and $\overset{\to }{\mathop{OB}}\,$.
Consider two points P and Q whose co – ordinates are,
$\overset{\to }{\mathop{P}}\,=\left( a,b,c \right)$ and $\overset{\to }{\mathop{Q}}\,=\left( x,y,z \right)$.
Then the value of PQ is, $\overset{\to }{\mathop{PQ}}\,=\left( \left( x-a \right)\overset{\to }{\mathop{i}}\,+\left( y-b \right)\overset{\to }{\mathop{j}}\,+\left( z-c \right)\overset{\to }{\mathop{k}}\, \right)$.
Now we have OA, where O is the origin then O = (0, 0, 0).
Using the above formula we have,
$OA=2\overset{\to }{\mathop{i}}\,+\overset{\to }{\mathop{j}}\,+\left( -2 \right)\overset{\to }{\mathop{k}}\,$ and similarly $OB=3\overset{\to }{\mathop{i}}\,-4\overset{\to }{\mathop{j}}\,+5\overset{\to }{\mathop{k}}\,$.
Now finally we will use the formula of dot product of two vectors which is given as,
$\overset{\to }{\mathop{P}}\,.\overset{\to }{\mathop{Q}}\,=\left| \overset{\to }{\mathop{P}}\, \right|\left| \overset{\to }{\mathop{Q}}\, \right|\cos \theta$, where $\left| \overset{\to }{\mathop{P}}\, \right|$ is the length of $\overset{\to }{\mathop{P}}\,$. $\left| \overset{\to }{\mathop{Q}}\, \right|$ is the length of $\overset{\to }{\mathop{Q}}\,$.
And $\theta$ is the angle between them as shown below,

The formula to compute length of $\overset{\to }{\mathop{P}}\,$ is,
$\left| \overset{\to }{\mathop{P}}\, \right|=\sqrt{{{\left( a \right)}^{2}}+{{\left( b \right)}^{2}}+{{\left( c \right)}^{2}}}$, where $\overset{\to }{\mathop{P}}\,=\left( a,b,c \right)$.
Now we will compute $\left| \overset{\to }{\mathop{OA}}\, \right|$ and $\left| \overset{\to }{\mathop{OB}}\, \right|$ using this formula,

& \therefore \left| \overset{\to }{\mathop{OA}}\, \right|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}+{{\left( -2 \right)}^{2}}} \\\ & \Rightarrow \left| \overset{\to }{\mathop{OA}}\, \right|=\sqrt{9}=3 \\\ \end{aligned}$$ Similarly, $$\left| \overset{\to }{\mathop{OB}}\, \right|=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( -4 \right)}^{2}}+{{5}^{2}}}$$ $$\begin{aligned} & \Rightarrow \left| \overset{\to }{\mathop{OB}}\, \right|=\sqrt{9+16+25} \\\ & \Rightarrow \left| \overset{\to }{\mathop{OB}}\, \right|=\sqrt{50} \\\ \end{aligned}$$ Finally we will use the formula of dot product stated above to find the angle ‘$$\theta $$’. Let $$\theta $$ be angle between OA & OB, then we have $$\overset{\to }{\mathop{OA}}\,.\overset{\to }{\mathop{OB}}\,=\left| \overset{\to }{\mathop{OA}}\, \right|\left| \overset{\to }{\mathop{OB}}\, \right|\cos \theta $$ Substituting values of $$\left| \overset{\to }{\mathop{OA}}\, \right|$$ & $$\left| \overset{\to }{\mathop{OB}}\, \right|$$ obtained above we have, $$\left( 2i+j+\left( -2 \right)k \right).\left( 3i-4j+5k \right)=3\times \sqrt{50}\cos \theta $$ Or product of two vectors given as, $$\left( ai+bj+ck \right).\left( xi+yj+zk \right)=ax+by+cz$$ Using this above we get, $$\begin{aligned} & \Rightarrow 6-4-10=3\sqrt{50}\times \cos \theta \\\ & \Rightarrow \cos \theta =\dfrac{6-14}{3\sqrt{50}} \\\ & \Rightarrow \cos \theta =\dfrac{-8}{5\times 3\sqrt{2}} \\\ \end{aligned}$$ Multiplying $$\sqrt{2}$$ on both numerator & denominator we get, $$\begin{aligned} & \Rightarrow \cos \theta =\dfrac{-8\sqrt{2}}{15\times 2} \\\ & \Rightarrow \cos \theta =\dfrac{-4\sqrt{2}}{15} \\\ \end{aligned}$$ Now because $$\overset{\to }{\mathop{OA}}\,$$ and $$\overset{\to }{\mathop{OB}}\,$$ are having obtuse angle in between so, $$\cos \theta $$ = +ve, anyway angle is always taken positive. $$\begin{aligned} & \Rightarrow \cos \theta =-\left( \dfrac{4\sqrt{2}}{15} \right) \\\ & \Rightarrow \cos \theta =\dfrac{4\sqrt{2}}{15} \\\ \end{aligned}$$ Taking $${{\cos }^{-1}}$$ both sides, $$\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{4\sqrt{2}}{15} \right)$$, which is option (a). **Note:** Do not get confuse while making $$\cos \theta $$ positive at the end of solution. If the student has any doubt then refer to the formula. $$\cos \theta =\left| \dfrac{\overset{\to }{\mathop{P}}\,.\overset{\to }{\mathop{Q}}\,}{\left| \overset{\to }{\mathop{P}}\, \right|\left| \overset{\to }{\mathop{Q}}\, \right|} \right|$$ for angle between P & Q as $$\theta $$. The angle is always considered positive. Note the key point to note in this question is that while calculating the position vector PQ where $$\overset{\to }{\mathop{P}}\,=\left( a,b,x \right)$$ and $$\overset{\to }{\mathop{Q}}\,=\left( x,y,z \right)$$ and $$\overset{\to }{\mathop{PQ}}\,=\left( x-a \right)\overset{\wedge }{\mathop{i}}\,+\left( y-b \right)\overset{\wedge }{\mathop{j}}\,+\left( z-c \right)\overset{\wedge }{\mathop{k}}\,$$. We can also use $$\overset{\to }{\mathop{QP}}\,=\overset{\to }{\mathop{PQ}}\,=\left( a-x \right)\overset{\wedge }{\mathop{i}}\,+\left( b-y \right)\overset{\wedge }{\mathop{j}}\,+\left( c-z \right)\overset{\wedge }{\mathop{k}}\,$$ this is valid until and unless you use this in whole solution calculations. Remember using $$\overset{\to }{\mathop{PQ}}\,=\left( x-a \right)\overset{\wedge }{\mathop{i}}\,+\left( y-b \right)\overset{\wedge }{\mathop{j}}\,+\left( z-c \right)\overset{\wedge }{\mathop{k}}\,$$ at one part and $$\overset{\to }{\mathop{QP}}\,=\left( a-x \right)\overset{\wedge }{\mathop{i}}\,+\left( b-y \right)\overset{\wedge }{\mathop{j}}\,+\left( c-z \right)\overset{\wedge }{\mathop{k}}\,$$ would differ the answer.