JEE MAIN Vector Previous Year Questions (PYQs) – Page 14 of 17

The line $L_1$ is parallel to the vector $\vec{a} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ and passes through the point $(7, 6, 2)$, and the line $L_2$ is parallel to the vector $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$ and passes through the point $(5, 3, 4)$. The shortest distance between the lines $L_1$ and $L_2$ is:

The vertices $B$ and $C$ of a $\triangle ABC$ lie on the line $\dfrac{x+2}{3}=\dfrac{y-1}{0}=\dfrac{z}{4}$ such that $BC=5$ units. Then the area (in sq. units) of this triangle, given that the point $A(1,-1,2)$, is:

Let $\vec a=2\hat i+\alpha\hat j+\hat k,\ \vec b=-\hat i+\hat k,\ \vec c=\beta\hat j-\hat k$, where $\alpha,\beta$ are integers and $\alpha\beta=-6$. Let the values of the ordered pair $(\alpha,\beta)$ for which the area of the parallelogram whose diagonals are $\vec a+\vec b$ and $\vec b+\vec c$ is $\dfrac{\sqrt{21}}{2}$ be $(\alpha_1,\beta_1)$ and $(\alpha_2,\beta_2)$. Then $\alpha_1^{,2}+\beta_1^{,2}-\alpha_2\beta_2$ is equal to:

Let $\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$, $\vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k}$, and a vector $\vec{c}$ be such that $(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{i} - 3\hat{j} + 12\hat{k}$ and $\vec{a} \cdot \vec{c} = 3$. If $\vec{b} \times \vec{c} = \vec{d}$, then $|\vec{a} \cdot \vec{d}|$ is equal to:

In a triangle $ABC$, right angled at the vertex $A$, if the position vectors of $A,B$ and $C$ are respectively $3\hat{i} + \hat{j} - \hat{k}$, $-\hat{i} + 3\hat{j} + p\hat{k}$ and $5\hat{i} + q\hat{j} - 4\hat{k}$, then the point $(p,q)$ lies on a line:

$ \text{Let A, B, C be three points whose position vectors respectively are } \vec{a} = \hat{i} + 4\hat{j} + 3\hat{k}, ; \vec{b} = 2\hat{i} + \alpha \hat{j} + 4\hat{k}, ; \alpha \in \mathbb{R}, ; \vec{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}. ; \text{If } \alpha \text{ is the smallest positive integer for which } \vec{a}, \vec{b}, \vec{c} \text{ are non-collinear, then the length of the median in } \triangle ABC \text{ through A is :}$

The shortest distance between the lines $\dfrac{x}{2} = \dfrac{y}{2} = \dfrac{z}{1}$ and $\dfrac{x + 2}{-1} = \dfrac{y - 4}{8} = \dfrac{z - 5}{4}$ lies in the interval:

Let a unit vector $\overrightarrow{OP}$ make angles $\alpha,\beta,\gamma$ with the positive directions of the coordinate axes $OX, OY, OZ$ respectively, where $\beta\in\left(0,\tfrac{\pi}{2}\right)$. If $\overrightarrow{OP}$ is perpendicular to the plane through points $(1,2,3)$, $(2,3,4)$ and $(1,5,7)$, then which one of the following is true?

Let $\vec a=-5\hat i+\hat j-3\hat k$, $\vec b=\hat i+2\hat j-4\hat k$ and $\vec c=\big(((\vec a\times\vec b)\times\hat i)\times\hat i\big)\times\hat i$. Then $\ \vec c\cdot(-\hat i+\hat j+\hat k)$ is equal to:

Let $L_1:\ \dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $L_2:\ \dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}$ be two lines. Which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?