Consider the planes 3x - 6y - 2z = 15 and 2x | vedclass

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(D) The normal vectors to the planes are $\vec{n_1} = 3\hat{i} - 6\hat{j} - 2\hat{k}$ and $\vec{n_2} = 2\hat{i} + \hat{j} - 2\hat{k}$.The direction ve

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$Statement-1$ is False,$Statement-2$ is True

Vedclass · Answer

(D) The normal vectors to the planes are $\vec{n_1} = 3\hat{i} - 6\hat{j} - 2\hat{k}$ and $\vec{n_2} = 2\hat{i} + \hat{j} - 2\hat{k}$.The direction vector $\vec{v}$ of the line of intersection is given by $\vec{n_1} 	imes \vec{n_2}$:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -6 & -2 \ 2 & 1 & -2 \end{vmatrix} = \hat{i}(12 - (-2)) - \hat{j}(-6 - (-4)) + \hat{k}(3 - (-12)) = 14\hat{i} + 2\hat{j} + 15\hat{k}$.Thus,$Statement-2$ is True.To find a point on the line,set $z = 0$ in the plane equations:$3x - 6y = 15 \Rightarrow x - 2y = 5$$2x + y = 5$Solving these,we get $x = 3, y = -1$. So,$(3, -1, 0)$ is a point on the line.The equation of the line is $\frac{x-3}{14} = \frac{y+1}{2} = \frac{z-0}{15} = t$.This gives $x = 14t + 3, y = 2t - 1, z = 15t$.Comparing this with $Statement-1$,the equations provided are $x = 3 + 14t, y = 1 + 2t, z = 15t$,which is incorrect because the $y$-coordinate is $2t - 1$,not $2t + 1$.Therefore,$Statement-1$ is False and $Statement-2$ is True.

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