Question
A vector of magnitude 4 which is equally inclined to the vectors $$\overrightarrow i + \overrightarrow j ,\,\overrightarrow j + \overrightarrow k $$ and $$\overrightarrow k + \overrightarrow i $$ is :
A.
$$\frac{4}{{\sqrt 3 }}\left( {\overrightarrow i - \overrightarrow j - \overrightarrow k } \right)$$
B.
$$\frac{4}{{\sqrt 3 }}\left( {\overrightarrow i + \overrightarrow j - \overrightarrow k } \right)$$
C.
$$\frac{4}{{\sqrt 3 }}\left( {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right)$$
D.
none of these
Answer :
$$\frac{4}{{\sqrt 3 }}\left( {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right)$$
Solution :
$$\eqalign{
& {\text{Let }}\overrightarrow a = x\overrightarrow i + y\overrightarrow j + z\overrightarrow k .{\text{ Then }}\left| {\overrightarrow a } \right| = 4 = \sqrt {{x^2} + {y^2} + {z^2}} \cr
& {\text{Now, }}\frac{{\overrightarrow a .\left( {\overrightarrow i + \overrightarrow j } \right)}}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow i + \overrightarrow j } \right|}} = \frac{{\overrightarrow a .\left( {\overrightarrow j + \overrightarrow k } \right)}}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow j + \overrightarrow k } \right|}} = \frac{{\overrightarrow a .\left( {\overrightarrow k + \overrightarrow i } \right)}}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow k + \overrightarrow i } \right|}}\left( {{\text{from the question}}} \right) \cr
& {\text{or }}x + y = y + z = z + x = t\,\,\left( {{\text{say}}} \right) \cr
& {\text{Adding, }}2\left( {x + y + z} \right) = 3t\,\,\,\,{\text{or}}\,\,x + y + z = \frac{{3t}}{2} \cr
& \therefore \,x = y = z = \frac{t}{2} \cr
& \therefore \,4 = \sqrt {{x^2} + {y^2} + {z^2}} \, \Rightarrow 16 = 3.{\left( {\frac{t}{2}} \right)^2} \cr
& \therefore t = \pm \frac{8}{{\sqrt 3 }}.{\text{ So, }}\overrightarrow a = \pm \frac{8}{{2\sqrt 3 }}\left( {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) \cr} $$