 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. $\dfrac{1}{3}i - \dfrac{4}{3}j + \dfrac{1}{3}k$

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.