Question

The vectors \[a,b{\text{ and }}c\] are equal in length and taken pairwise, they make equal angles. If \[a = i + j,b = j + k{\text{ and }}c\] makes an obtuse angle with the base vector \[i\]and \[c\] is equal to
A. \[i + k\]
B. \[ - i + 4j - k\]
C. \[ - \dfrac{1}{3}i + \dfrac{4}{3}j - \dfrac{1}{3}k\]
D. \[\dfrac{1}{3}i - \dfrac{4}{3}j + \dfrac{1}{3}k\]

Answer 1

Hint: First of all, find the length of vector \[c\].Then find the angles made by the vectors \[a,b{\text{ and }}c\] taken pair wise to get the equations in terms of components of vector \[c\]. So, use this concept to reach the solution of the given problem.

Complete step-by-step solution -
The length of the vector \[r = xi + yj + zk\] is given by \[\left| r \right| = \sqrt {{x^2} + {y^2} + {z^2}} \].
The length of vector \[a = i + j\] is \[\left| a \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt 2 \]
The length of vector \[b = j + k\] is \[\left| b \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt 2 \]
Since three vectors have equal lengths \[\left| c \right| = \sqrt 2 \]
Let vector \[c = {c_1}i + {c_2}j + {c_3}k\]
Since vector \[c\]makes an obtuse angle with \[i\], then the dot product between them is less than zero i.e., \[c.i = {c_1} < 0\]
We know that the angle between the vectors \[x{\text{ and }}y\] is given by \[\theta = {\cos ^{ - 1}}\dfrac{{x.y}}{{\left| x \right|\left| y \right|}}\]
Also given that the angle between the vectors are equal. So, we have
\[{\cos ^{ - 1}}\dfrac{{a.b}}{{\left| a \right|\left| b \right|}} = {\cos ^{ - 1}}\dfrac{{a.c}}{{\left| a \right|\left| c \right|}} = {\cos ^{ - 1}}\dfrac{{b.c}}{{\left| b \right|\left| c \right|}}\]
Now consider
\[
   \Rightarrow a.b = \left( {i + j} \right).\left( {j + k} \right) = {j^2} = 1 \\
   \Rightarrow a.c = \left( {i + j} \right).\left( {{c_1}i + {c_2}j + {c_3}k} \right) = {c_1} + {c_2} \\
   \Rightarrow b.c = \left( {j + k} \right).\left( {{c_1}i + {c_2}j + {c_3}k} \right) = {c_2} + {c_3} \\
\]
Taking \[{\cos ^{ - 1}}\dfrac{{a.b}}{{\left| a \right|\left| b \right|}} = {\cos ^{ - 1}}\dfrac{{a.c}}{{\left| a \right|\left| c \right|}}\], we have
\[
   \Rightarrow {\cos ^{ - 1}}\dfrac{1}{{\sqrt 2 \sqrt 2 }} = {\cos ^{ - 1}}\dfrac{{{c_1} + {c_2}}}{{\sqrt 2 \sqrt 2 }} \\
   \Rightarrow \dfrac{1}{{\sqrt 2 \sqrt 2 }} = \dfrac{{{c_1} + {c_2}}}{{\sqrt 2 \sqrt 2 }} \\
   \Rightarrow \dfrac{1}{2} = \dfrac{{{c_1} + {c_2}}}{2} \\
   \Rightarrow {c_1} + {c_2} = 1 \\
  \therefore {c_2} = 1 - {c_1}..................................................\left( 1 \right) \\
\]
Taking \[{\cos ^{ - 1}}\dfrac{{a.b}}{{\left| a \right|\left| b \right|}} = {\cos ^{ - 1}}\dfrac{{b.c}}{{\left| b \right|\left| c \right|}}\]
\[
   \Rightarrow {\cos ^{ - 1}}\dfrac{1}{{\sqrt 2 \sqrt 2 }} = {\cos ^{ - 1}}\dfrac{{{c_2} + {c_3}}}{{\sqrt 2 \sqrt 2 }} \\
   \Rightarrow \dfrac{1}{{\sqrt 2 \sqrt 2 }} = \dfrac{{{c_2} + {c_3}}}{{\sqrt 2 \sqrt 2 }} \\
   \Rightarrow \dfrac{1}{2} = \dfrac{{{c_2} + {c_3}}}{2} \\
   \Rightarrow {c_2} + {c_3} = 1 \\
\]
From equation (1) we have
\[
   \Rightarrow 1 - {c_1} + {c_3} = 1 \\
  \therefore {c_3} = {c_1}.................................\left( 2 \right) \\
\]
Since \[\left| c \right| = \sqrt 2 \]
\[ \Rightarrow \sqrt {{{\left( {{c_1}} \right)}^2} + {{\left( {{c_2}} \right)}^2} + {{\left( {{c_3}} \right)}^2}} = \sqrt 2 \]
Squaring on both sides we get
\[ \Rightarrow {\left( {{c_1}} \right)^2} + {\left( {{c_2}} \right)^2} + {\left( {{c_3}} \right)^2} = 2\]
From equation (1) and (2) we het
\[
   \Rightarrow {\left( {{c_1}} \right)^2} + {\left( {1 - {c_1}} \right)^2} + {\left( {{c_1}} \right)^2} = 2 \\
   \Rightarrow {c_1}^2 + 1 - 2{c_1} + {c_1}^2 + {c_1}^2 = 2 \\
   \Rightarrow 3{c_1}^2 - 2{c_1} = 2 - 1 = 1 \\
   \Rightarrow 3{c_1}^2 - 3{c_1} + {c_1} - 1 = 0 \\
\]
Taking the common terms, we have
\[
   \Rightarrow 3{c_1}\left( {{c_1} - 1} \right) + 1\left( {{c_1} - 1} \right) = 0 \\
   \Rightarrow \left( {3{c_1} + 1} \right)\left( {{c_1} - 1} \right) = 0 \\
  \therefore {c_1} = 1, - \dfrac{1}{3} \\
\]
Since \[{c_1} < 0\]
The value of \[{c_1}\] is \[ - \dfrac{1}{3}\]
From equation (1) we have
\[
   \Rightarrow {c_2} = 1 - \left( { - \dfrac{1}{3}} \right) = 1 + \dfrac{1}{3} \\
  \therefore {c_2} = \dfrac{4}{3} \\
\]
From equation (2) we have
\[
   \Rightarrow {c_3} = {c_1} = - \dfrac{1}{3} \\
  \therefore {c_3} = - \dfrac{1}{3} \\
\]
Hence vector \[c = {c_1}i + {c_2}j + {c_3}k\] is \[c = - \dfrac{1}{3}i + \dfrac{4}{3}j - \dfrac{1}{3}k\]
Thus, the correct option is C. \[ - \dfrac{1}{3}i + \dfrac{4}{3}j - \dfrac{1}{3}k\]

Note: The length of the vector \[r = xi + yj + zk\] is given by \[\left| r \right| = \sqrt {{x^2} + {y^2} + {z^2}} \]. The angle between the vectors \[x{\text{ and }}y\] is given by \[\theta = {\cos ^{ - 1}}\dfrac{{x.y}}{{\left| x \right|\left| y \right|}}\]. The angle made by the two vectors is said to be an obtuse angle when their dot product is less than zero.