If the angle between the vectors i + k and i-j + ak is π3, then a =

a=P+K b=i-j+ak cosπ3=a.bab=12=1+a1+11+1+a2 =(2+a2)=4(1+a)2 =2+a2=2+2a2+4a =a2+4a=0 a=-4,a=0

If the angle between the vectors i + k and i-j + ak is π3, then a =

a=P+K b=i-j+ak cosπ3=a.bab=12=1+a1+11+1+a2 =(2+a2)=4(1+a)2 =2+a2=2+2a2+4a =a2+4a=0 a=-4,a=0

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If the angle between the vectors $i + k$ and $i - j + a k$ is $\frac{π}{3}$ , then $a =$

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$a = P + K b = i - j + ak \cos \frac{π}{3} = \frac{a . b}{|a| |b|} = \frac{1}{2} = \frac{1 + a}{\sqrt{1 + 1} \sqrt{1 + 1 + a^{2}}} = (2 + a^{2}) = 4 {(1 + a)}^{2} = 2 + a^{2} = 2 + 2 a^{2} + 4 a = a^{2} + 4 a = 0 a = - 4, a = 0$

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If the angle between the vectors $i + k$ and $i - j + a k$ is $\frac{π}{3}$ , then $a =$