JEE MAIN Vector Previous Year Questions (PYQs) – Page 8 of 17

Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$
and $\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the
points $(0, 0), (\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is

For any vector $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$, with $10|a_i| < 1, \; i = 1, 2, 3$, consider the following statements: (A): $\max \{|a_1|, |a_2|, |a_3|\} \le |\vec{a}|$ (B): $|\vec{a}| \le 3 \max \{|a_1|, |a_2|, |a_3|\}$

Let $\overrightarrow a = \widehat i + \widehat j + 2\widehat k$, $\overrightarrow b = 2\widehat i - 3\widehat j + \widehat k$ and $\overrightarrow c = \widehat i - \widehat j + \widehat k$ be three given vectors. Let $\overrightarrow v $ be a vector in the plane of $\overrightarrow a $ and $\overrightarrow b $ whose projection on $\overrightarrow c $ is ${2 \over {\sqrt 3 }}$. If $\overrightarrow v \,.\,\widehat j = 7$, then $\overrightarrow v \,.\,\left( {\widehat i + \widehat k} \right)$ is equal to :

Let $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$ and $\vec b=b_1\hat i+b_2\hat j+b_3\hat k$ be two vectors such that $|\vec a|=1,\ \vec a\cdot\vec b=2$ and $|\vec b|=4$. If $\vec c=2(\vec a\times\vec b)-3\vec b$, then the angle between $\vec b$ and $\vec c$ is:

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\overrightarrow{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}$. If $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{r}$, $\overrightarrow{r} \cdot (\alpha \hat{i} + 2\hat{j} + \hat{k}) = 3$ and $\overrightarrow{r} \cdot (2\hat{i} + 5\hat{j} - \alpha \hat{k}) = -1$, $\alpha \in \mathbb{R}$, then the value of $\alpha + |\overrightarrow{r}|^2$ is equal to :

The set of all $\alpha$ for which the vectors $\vec a=\alpha t,\hat i+6,\hat j-3,\hat k$ and $\vec b=t,\hat i-2,\hat j-2\alpha t,\hat k$ are inclined at an obtuse angle for all $t\in\mathbb{R}$ is:

Let $\vec{a}=2\hat{i}+\hat{j}-2\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}$. Let $\vec{c}$ be a vector such that $|\vec{c}-\vec{a}|=3$, $|(\vec{a}\times\vec{b})\times\vec{c}|=3$ and the angle between $\vec{c}$ and $\vec{a}\times\vec{b}$ is $30^\circ$. Then $\vec{a}\cdot\vec{c}$ is equal to :

Let $\vec{a}$ be a non-zero vector parallel to the line of intersection of the two planes described by $\hat{i} + \hat{j}, \; \hat{i} + \hat{k}$ and $\hat{i} - \hat{j}, \; \hat{j} - \hat{k}$. If $\theta$ is the angle between the vector $\vec{a}$ and the vector $\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{a} \cdot \vec{b} = 6$, then the ordered pair $(\theta, |\vec{a} \times \vec{b}|)$ is equal to:

Let $(\alpha,\beta,\gamma)$ be the foot of the perpendicular from the point $(1,2,3)$ on the line \[ \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}. \] Then $19(\alpha+\beta+\gamma)$ is equal to:

The length of the perpendicular from the point $(2,-1,4)$ on the straight line $\displaystyle \frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}$ is: