JEE Main 2003 — Mathematics Vectors & 3D Geometry

Let the line $\mathrm{L}_{1}$ be parallel to the vector $-3 \hat{i}+2 \hat{j}+4 \hat{k}$ and pass through the point (2,6,7), and the line $\mathrm{L}_{2}$ be parallel to the vector $2 \hat{i}+\hat{j}+3 \hat{k}$ and pass through the point $(4,3,5)$. If the line $\mathrm{L}_{3}$ is parallel to the vector $-3 \hat{i}+5 \hat{j}+16 \hat{k}$ and intersects the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ at the points C and D, respectively, then $|\overrightarrow{C D}|^{2}$ is equal to :

Let the image of the point $P(0, -5, 0)$ in the line $\dfrac{x-1}{2} = \dfrac{y}{1} = \dfrac{z+1}{-2}$ be the point $R$ and the image of the point $Q\left(0, \dfrac{-1}{2}, 0\right)$ in the line $\dfrac{x-1}{-1} = \dfrac{y+9}{4} = \dfrac{z+1}{1}$ be the point $S$. Then the square of the area of the parallelogram $PQRS$ is __________.

The square of the distance of the point $P(5, 6, 7)$ from the line $\dfrac{x-2}{2} = \dfrac{y-5}{3} = \dfrac{z-2}{4}$ is equal to:

Let $\overrightarrow{\mathrm{AB}}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\overrightarrow{\mathrm{AD}}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \lambda \in \mathbb{R}$. Let the projection of the vector $\vec{v}=\hat{i}+\hat{j}+\hat{k}$ on the diagonal $\overrightarrow{\mathrm{AC}}$ of the parallelogram ABCD be of length one unit. If $\alpha, \beta$, where $\alpha>\beta$, be the roots of the equation $\lambda^{2} x^{2}-6 \lambda x+5=0$, then $2 \alpha-\beta$ is equal to

Let $\vec{a}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\vec{c}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to