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# If $\vec{a}=2\hat{i}+2\hat{j}+3\hat{k},\vec{b}=-\hat{i}+2\hat{j}+\hat{k}$and $\vec{c}=3\hat{i}+\hat{j}$ such that $\vec{a}+\lambda \vec{b}$ is perpendicular to the vector $\vec{c}$ then find the value of $\lambda$.

Last updated date: 17th Mar 2023
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Hint: To solve the question, we have to apply the properties of the dot product of vectors to calculate the value of $\lambda$.

Complete step-by-step Solution:
The vector $\vec{a}+\lambda \vec{b}$ is given that it is perpendicular to the vector $\vec{c}$.
The given values of the vectors $\vec{a},\vec{b},\vec{c}$ are $2\hat{i}+2\hat{j}+3\hat{k}$ , $-\hat{i}+2\hat{j}+\hat{k}$, $3\hat{i}+\hat{j}$respectively.
By substituting the value of vector $\vec{a},\vec{b}$ in $\vec{a}+\lambda \vec{b}$ we get,
$\vec{a}+\lambda \vec{b}=2\hat{i}+2\hat{j}+3\hat{k}+\lambda \left( -\hat{i}+2\hat{j}+\hat{k} \right)$
$=\left( 2-\lambda \right)\hat{i}+\left( 2+2\lambda \right)\hat{j}+\left( 3+\lambda \right)\hat{k}$
We know that a dot product of $\vec{a},\vec{b}$is given the formula $\vec{a}.\vec{b}=\left\| {\vec{a}} \right\|\left\| {\vec{b}} \right\|\cos \theta$
Where $\theta ,\left\| {} \right\|$ represent the angle between the given vectors and the magnitude of the vector respectively.
Now we know that when $\theta ={{90}^{0}}$, $\cos \left( {{90}^{0}} \right)=0$ thus the dot product of perpendicular vectors is 0.
Thus, the product of the vectors $\vec{a}+\lambda \vec{b}$ and $\vec{c}$is equal to 0.
$\left( \vec{a}+\lambda \vec{b} \right).\vec{c}=\left( \left( 2-\lambda \right)\hat{i}+\left( 2+2\lambda \right)\hat{j}+\left( 3+\lambda \right)\hat{k} \right).\left( 3\hat{i}+\hat{j} \right)$
$=\left( 3\left( 2-\lambda \right)\hat{i}.\hat{i}+3\left( 2+2\lambda \right)\hat{j}.\hat{i}+3\left( 3+\lambda \right)\hat{k}.\hat{i} \right)+\left( \left( 2-\lambda \right)\hat{i}.\hat{j}+\left( 2+2\lambda \right)\hat{j}.\hat{j}+\left( 3+\lambda \right)\hat{k}.\hat{j} \right)$ ….. (1)
We know that the dot product of the orthogonal vectors $\hat{i},\hat{j},\hat{k}$ is given by the following values.
$\hat{i}.\hat{j}=\hat{j}.\hat{k}=\hat{i}.\hat{k}=0$ and $\hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1$
By substituting the values in equation (1) and equating it the value of 0 since $\left( \vec{a}+\lambda \vec{b} \right).\vec{c}=0$we get,
$\left( 3\left( 2-\lambda \right)\times 1+3\left( 2+2\lambda \right)\times 0+3\left( 3+\lambda \right)\times 0 \right)+\left( \left( 2-\lambda \right)\times 0+\left( 2+2\lambda \right)\times 1+\left( 3+\lambda \right)\times 0 \right)=0$
$3\left( 2-\lambda \right)+\left( 2+2\lambda \right)=0$
$6-3\lambda +2+2\lambda =0$
$8-\lambda =0$
$\lambda =8$
Thus, for the given condition $\vec{a}+\lambda \vec{b}$ is perpendicular to the vector $\vec{c}$, the value of $\lambda$is equal to 8.

Note: The possibility of mistake can be not able to analyse that dot product can be used for solving the given problem. The other possibility of mistake is doing simple multiplication using a vector product, the dot product of vectors is different from normal scalar multiplication. The alternative way of solving the question can be applying the properties of cross product of vectors. The cross product of perpendicular vectors is equal to 1 and the cross-product property is $\hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0$.