Question

If $\mathop a\limits^ \to = 2i + 2f + 3k,\mathop b\limits^ \to = - i + 2j + k\& \mathop c\limits^ \to = 3i + j$ are such that $\mathop a\limits^ \to + \lambda \mathop b\limits^ \to$ is perpendicular to $\mathop c\limits^ \to$ then find the value of $\lambda$ .

Answer 1

It is given that the $\mathop a\limits^ \to + \lambda \mathop b\limits^ \to$ and $\mathop c\limits^ \to$ are perpendicular to each other which means that the dot product of both the vectors will be zero.

Complete step by step answer:
As $\mathop A\limits^ \to \& \mathop B\limits^ \to$ be any two vectors then there dot product is represented by $\mathop A\limits^ \to .\mathop B\limits^ \to = \left| A \right|\left| B \right|\cos \theta$
Where $\theta$ is the angle between the two vectors
If the two vectors are mutually perpendicular then $\theta = {90^ \circ }$
Putting this we will get,

\Rightarrow \mathop A\limits^ \to .\mathop B\limits^ \to = \left| A \right|\left| B \right|\cos {90^ \circ }\\\ \mathop { \Rightarrow A}\limits^ \to .\mathop B\limits^ \to = \left| A \right|\left| B \right| \times 0\\\ \Rightarrow \mathop A\limits^ \to .\mathop B\limits^ \to = 0 \end{array}$$ Using this result we can get As the vectors are $$\mathop a\limits^ \to = 2i + 2f + 3k,\mathop b\limits^ \to = - i + 2j + k\& \mathop c\limits^ \to = 3i + j$$ Then $$\mathop a\limits^ \to + \lambda \mathop b\limits^ \to $$ is $$\begin{array}{l} \mathop {\therefore {\rm{ }}a}\limits^ \to + \lambda \mathop {{\rm{ }}b}\limits^ \to = (2i + 2j + 3k) + \lambda ( - i + 2j + k)\\\ \Rightarrow \mathop {{\rm{ }}a}\limits^ \to + \lambda \mathop {{\rm{ }}b}\limits^ \to = (2 - \lambda )i + (2 + 2\lambda )j + (3 + \lambda )k \end{array}$$ Then $$\begin{array}{l} \therefore \left( {\mathop {{\rm{ }}a}\limits^ \to + \lambda \mathop {{\rm{ }}b}\limits^ \to } \right).\mathop c\limits^ \to = 0\\\ \Rightarrow \left[ {(2 - \lambda )i + (2 + 2\lambda )j + (3 + \lambda )k} \right].\left[ {3i + j} \right] = 0\\\ \Rightarrow (2 - \lambda )3 + (2 + 2\lambda )1 + (3 + \lambda ) = 0\\\ \Rightarrow 6 - 3\lambda + 2 + 2\lambda = 0\\\ \Rightarrow - \lambda + 8 = 0\\\ \Rightarrow \lambda = 8 \end{array}$$ Which means that the value of $$\lambda $$ is 8 **Note:** While doing dot product of vectors we must keep in mind that the units of i will only multiply with the units i and the same goes for the rest as it has be obtained that when we do dot product of any other component with i we only get 0 and the same goes for the rest of the components.