A straight line is given by vec{r} = (1 + t)h | vedclass

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(A) The given equation of the line is $\vec{r} = (\hat{i} + \hat{k}) + t(\hat{i} + 3\hat{j} - \hat{k})$,where $t \in R$.Since the line lies in the pla

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$9$

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(A) The given equation of the line is $\vec{r} = (\hat{i} + \hat{k}) + t(\hat{i} + 3\hat{j} - \hat{k})$,where $t \in R$.Since the line lies in the plane $x + y + cz = d$,every point on the line must satisfy the plane equation.The point $(1, 0, 1)$ lies on the line (at $t = 0$). Substituting this into the plane equation:$1 + 0 + c(1) = d \Rightarrow 1 + c = d$  .....$(1)$Also,the direction vector of the line $\vec{v} = \hat{i} + 3\hat{j} - \hat{k}$ must be perpendicular to the normal vector of the plane $\vec{n} = \hat{i} + \hat{j} + c\hat{k}$.Therefore,their dot product must be zero:$(1)(1) + (3)(1) + (-1)(c) = 0$$1 + 3 - c = 0 \Rightarrow c = 4$.Substituting $c = 4$ into equation $(1)$:$1 + 4 = d \Rightarrow d = 5$.Thus,the value of $(c + d) = 4 + 5 = 9$.

$A$ straight line is given by $\vec{r} = (1 + t)\hat{i} + 3t\hat{j} + (1 - t)\hat{k}$ where $t \in R$. If this line lies in the plane $x + y + cz = d$,then the value of $(c + d)$ is

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