If hat{i}+hat{j}, hat{j}+hat{k}, hat{k}+hat{i | vedclass

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(B) The position vectors are $\vec{A} = \hat{i}+\hat{j}$, $\vec{B} = \hat{j}+\hat{k}$, $\vec{C} = \hat{k}+\hat{i}$, $\vec{D} = \hat{i}-\hat{j}$, $\vec

Vedclass · Accepted Answer

$\frac{1}{2} \hat{i}+\hat{j}+\frac{1}{2} \hat{k}$

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(B) The position vectors are $\vec{A} = \hat{i}+\hat{j}$, $\vec{B} = \hat{j}+\hat{k}$, $\vec{C} = \hat{k}+\hat{i}$, $\vec{D} = \hat{i}-\hat{j}$, $\vec{E} = \hat{j}-\hat{k}$.Equation of line $AB$ passing through $\vec{A}$ and $\vec{B}$ is $\vec{r} = \vec{A} + \lambda(\vec{B}-\vec{A}) = (\hat{i}+\hat{j}) + \lambda(-\hat{i}+\hat{k})$.So, $x = 1-\lambda, y = 1, z = \lambda$.The plane passing through $C, D, E$ has the normal vector $\vec{n} = (\vec{D}-\vec{C}) 	imes (\vec{E}-\vec{C})$.$\vec{D}-\vec{C} = \hat{i}-\hat{j}-\hat{k}-\hat{i} = -\hat{j}-\hat{k}$.$\vec{E}-\vec{C} = \hat{j}-\hat{k}-\hat{k}-\hat{i} = -\hat{i}+\hat{j}-2\hat{k}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -1 & -1 \ -1 & 1 & -2 \end{vmatrix} = \hat{i}(2+1) - \hat{j}(0-1) + \hat{k}(0-1) = 3\hat{i} + \hat{j} - \hat{k}$.The equation of the plane is $3(x-0) + 1(y-1) - 1(z-1) = 0 \Rightarrow 3x + y - z = 0$.Substituting the line coordinates into the plane equation: $3(1-\lambda) + 1 - \lambda = 0 \Rightarrow 3 - 3\lambda + 1 - \lambda = 0 \Rightarrow 4 = 4\lambda \Rightarrow \lambda = \frac{1}{2}$.Thus, the point of intersection is $(1-\frac{1}{2})\hat{i} + 1\hat{j} + \frac{1}{2}\hat{k} = \frac{1}{2}\hat{i} + \hat{j} + \frac{1}{2}\hat{k}$.

If $\hat{i}+\hat{j}, \hat{j}+\hat{k}, \hat{k}+\hat{i}, \hat{i}-\hat{j}, \hat{j}-\hat{k}$ are the position vectors of the points $A, B, C, D, E$ respectively, then the point of intersection of the line $AB$ and the plane passing through $C, D, E$ is

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