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

Hint: 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,
\[\begin{align}
  & \therefore \left| \overset{\to }{\mathop{OA}}\, \right|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}+{{\left( -2 \right)}^{2}}} \\
 & \Rightarrow \left| \overset{\to }{\mathop{OA}}\, \right|=\sqrt{9}=3 \\
\end{align}\]
Similarly, \[\left| \overset{\to }{\mathop{OB}}\, \right|=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( -4 \right)}^{2}}+{{5}^{2}}}\]
\[\begin{align}
  & \Rightarrow \left| \overset{\to }{\mathop{OB}}\, \right|=\sqrt{9+16+25} \\
 & \Rightarrow \left| \overset{\to }{\mathop{OB}}\, \right|=\sqrt{50} \\
\end{align}\]
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,
\[\begin{align}
  & \Rightarrow 6-4-10=3\sqrt{50}\times \cos \theta \\
 & \Rightarrow \cos \theta =\dfrac{6-14}{3\sqrt{50}} \\
 & \Rightarrow \cos \theta =\dfrac{-8}{5\times 3\sqrt{2}} \\
\end{align}\]
Multiplying \[\sqrt{2}\] on both numerator & denominator we get,
\[\begin{align}
  & \Rightarrow \cos \theta =\dfrac{-8\sqrt{2}}{15\times 2} \\
 & \Rightarrow \cos \theta =\dfrac{-4\sqrt{2}}{15} \\
\end{align}\]
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.
\[\begin{align}
  & \Rightarrow \cos \theta =-\left( \dfrac{4\sqrt{2}}{15} \right) \\
 & \Rightarrow \cos \theta =\dfrac{4\sqrt{2}}{15} \\
\end{align}\]
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.