Practicing Success
$\vec a$ and $\vec c$ are unit collinear vectors and $|\vec b|= 6$, then $\vec b-3\vec c=λ\vec a$, if $λ$ is |
-9, 3 9, 3 3, -3 none of these |
-9, 3 |
We have, $\vec b-3\vec c=λ\vec a$ Taking scalar product with $\vec c$, we have $(\vec b-3\vec c). \vec c =λ(\vec a.\vec c)$ $⇒\vec b.\vec c-3(\vec c.\vec c)=λ(\vec a.\vec c)$ $⇒\vec b.\vec c-3=λ$ [$∵|\vec a|=|\vec c|=1$ and $\vec a$ and $\vec c$ are collinear vectors] $⇒\vec b.\vec c=3+λ$ Again, $\vec b-3\vec c=λ\vec a$ $⇒|\vec b-3\vec c|=|λ\vec a|$ $⇒|\vec b-3\vec c|^2=λ^2|\vec a|^2$ $⇒|\vec b|^2+9|\vec c|^2-6(\vec b.\vec c)=λ^2|\vec a|^2$ $⇒36+9-6(3+λ)=λ^2$ [Using (i)] $⇒27-6λ=λ^2⇒λ^2+6λ-27=0⇒λ=-9,3$ |