Question

If the angle between two vectors i+k and i-j+ak is
π/3, then the value of a is
(a) 2
(b)4
(c)-2
(d)0​

Answer 1

Answer:

1. 4 is the write answer hope it help you please mark me brilliant htffgg11123943375647439393494

Answer 2

Given,

Vectors,

$\[\hat i + \hat k\]$ and $\[\hat i - \hat j + a\hat k\]$

Solution,

Consider the angle between the vectors $\[a\hat i + b\hat j + c\hat k\]$ and $\[l\hat i + m\hat j + n\hat k\]$ is $\[\theta \]$.

$\[\cos \theta = \frac{{\left( {a\hat i + b\hat j + c\hat k} \right).\left( {l\hat i + m\hat j + n\hat k} \right)}}{{\sqrt {\left( {{a^2} + {b^2} + {c^2}} \right)} \sqrt {\left( {{l^2} + {m^2} + {n^2}} \right)} }}\]$

$\[ \Rightarrow \cos \left( {\frac{\pi }{3}} \right) = \frac{{\left( {\hat i + \hat k} \right).\left( {\hat i - \hat j + a\hat k} \right)}}{{\sqrt {\left( {{1^2} + {1^2}} \right)} \sqrt {\left( {{1^2} + {{\left( { - 1} \right)}^2} + {a^2}} \right)} }}\]$

$\[ \Rightarrow \cos 60^\circ = \frac{{1 + a}}{{\sqrt 2 \sqrt {\left( {2 + {a^2}} \right)} }}\]$

$\[ \Rightarrow \frac{1}{2} = \frac{{1 + a}}{{\sqrt 2 \sqrt {\left( {2 + {a^2}} \right)} }}\]$

Squaring on both side,

$\[ \Rightarrow \frac{1}{2} = \frac{{{{\left( {1 + a} \right)}^2}}}{{\left( {2 + {a^2}} \right)}}\]$

$\[ \Rightarrow 2\left( {1 + {a^2} + 2a} \right) = 2 + {a^2}\]$

$\[ \Rightarrow 2 + 2{a^2} + 4a = 2 + {a^2}\]$

$\[ \Rightarrow {a^2} + 4a = 0\]$

$\[ \Rightarrow a\left( {a + 4} \right) = 0\]$

$\[ \Rightarrow a = 0, - 4\]$

Hence the value of a is $\[0\]$ or $\[ - 4\]$

Hence the correct option is (d), i.e. 0