Vectors & 3D Geometry PYQ — Page 4

If the equation of the line passing through the point $\left(0,-\frac{1}{2}, 0\right)$ and perpendicular to the lines $\vec{r}=\lambda(\hat{i}+a \hat{j}+b \hat{k})$ and $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}-6 \hat{\mathrm{k}})+\mu(-b \hat{\mathrm{i}}+\mathrm{a} \hat{\mathrm{j}}+5 \hat{\mathrm{k}})$ is $\frac{\mathrm{x}-1}{-2}=\frac{\mathrm{y}+4}{\mathrm{~d}}=\frac{\mathrm{z}-\mathrm{c}}{-4}$, then $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}$ is equal to :

Let \(\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\quad \overrightarrow{\mathrm{b}}=2 \hat{i}+7 \hat{j}+3 \hat{k} . \quad\) Let \(\mathrm{L}_1: \overrightarrow{\mathrm{r}}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda \overrightarrow{\mathrm{a}}, \lambda \in \mathbf{R}\) and \(\mathrm{L}_2: \overrightarrow{\mathrm{r}}=(\hat{j}+\hat{k})+\mu \overrightarrow{\mathrm{b}}, \mu \in \mathbf{R}\) be two lines. If the line \(\mathrm{L}_3\) passes through the point of intersection of \(\mathrm{L}_1\) and \(L_2\), and is parallel to \(\vec{a}+\vec{b}\), then \(L_3\) passes through the point :

Let $\hat{a}$ be a unit vector perpendicular to the vectors $\overrightarrow{\mathrm{b}}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\overrightarrow{\mathrm{c}}=2 \hat{i}+3 \hat{j}-\hat{k}$, and makes an angle of $\cos ^{-1}\left(-\frac{1}{3}\right)$ with the vector $\hat{i}+\hat{j}+\hat{k}$. If $\hat{\mathrm{a}}$ makes an angle of $\frac{\pi}{3}$ with the vector $\hat{i}+\alpha \hat{j}+\hat{k}$, then the value of $\alpha$ is :

Let the area of the triangle formed by the lines $x+2=y-1=z, \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}$ and $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}$ be $A$. Then $A^2$ is equal to ________

Let $\mathrm{L}_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $\mathrm{L}_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in \mathbf{R}$, be two lines, which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$, then the value of $26 \alpha(\mathrm{~PB})^2$ is _________

Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$. Then $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$ is :

If $\vec{a}$ is nonzero vector such that its projections on the vectors $2 \hat{i}-\hat{j}+2 \hat{k}, \hat{i}+2 \hat{j}-2 \hat{k}$ and $\hat{k}$ are equal, then a unit vector along $\vec{a}$ is:

If the components of $\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\overrightarrow{\mathrm{b}}=3 \hat{i}+\hat{j}-\hat{k}$ respectively, are $\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})$, then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Let $\vec{c}$ be the projection vector of $\vec{b}=\lambda \hat{i}+4 \hat{k}, \lambda\gt0$, on the vector $\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k}$. If $|\vec{a}+\vec{c}|=7$, then the area of the parallelogram formed by the vectors $\vec{b}$ and $\vec{c}$ is ________

Let the distance between two parallel lines be 5 units and a point $P$ lie between the lines at a unit distance from one of them. An equilateral triangle $P Q R$ is formed such that $Q$ lies on one of the parallel lines, while $R$ lies on the other. Then $(Q R)^2$ is equal to _______ -.

Let a line passing through the point $(4,1,0)$ intersect the line $L_1 ; \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $\mathrm{A} \quad(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then $\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ \mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right|$ is equal to

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 :

Let P be the foot of the perpendicular from the point $(1,2,2)$ on the line $\mathrm{L}: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$. Let the line $\vec{r}=(-\hat{i}+\hat{j}-2 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}), \lambda \in \mathbf{R}$, intersect the line L at Q . Then $2(\mathrm{PQ})^2$ is equal to :

Let in a $\triangle A B C$, the length of the side $A C$ be 6 , the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle \mathrm{ABC}$ is:

Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right angled triangle $P Q R$, where $R$ is the point $(3,-2,1)$, is

The position vectors of points A and B are 2i+3j+k and 4i+j-2k. The position vector of the midpoint of AB is:

Let the position vectors of three vertices of a triangle be $4 \vec{p}+\vec{q}-3 \vec{r},-5 \vec{p}+\vec{q}+2 \vec{r}$ and $2 \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}+2 \overrightarrow{\mathrm{r}}$. If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p}+\vec{q}+\vec{r}}{4}$ and $\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r}$ respectively, then $\alpha+2 \beta+5 \gamma$ is equal to :

If the square of the shortest distance between the lines $\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}$ and $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}$ is $\frac{\mathrm{m}}{\mathrm{n}}$, where $\mathrm{m}, \mathrm{n}$ are coprime numbers, then $\mathrm{m}+\mathrm{n}$ is equal to :

Let $\vec{a}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{d}}=\vec{a} \times \overrightarrow{\mathrm{b}}$. If $\overrightarrow{\mathrm{c}}$ is a vector such that $\vec{a} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|$, $|\overrightarrow{\mathrm{c}}-2 \vec{a}|^2=8$ and the angle between $\overrightarrow{\mathrm{d}}$ and $\overrightarrow{\mathrm{c}}$ is $\frac{\pi}{4}$, then $|10-3 \overrightarrow{\mathrm{~b}} \cdot \overrightarrow{\mathrm{c}}|+|\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{c}}|^2$ is equal to

Let the angle $\theta, 0 \lt \theta \lt \frac{\pi}{2}$ between two unit vectors $\hat{\mathrm{a}}$ and $\hat{\mathrm{b}}$ be $\sin ^{-1}\left(\frac{\sqrt{65}}{9}\right)$. If the vector $\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{a}}+6 \hat{\mathrm{~b}}+9(\hat{\mathrm{a}} \times \hat{\mathrm{b}}),$ then the value of $9(\overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{a}})-3(\overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{b}})$ is

Let \(\overrightarrow{\mathrm{a}}=2 \hat{i}-\hat{j}+3 \hat{k}, \overrightarrow{\mathrm{~b}}=3 \hat{i}-5 \hat{j}+\hat{k}\) and \(\overrightarrow{\mathrm{c}}\) be a vector such that \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}\) and \((\vec{a}+\vec{c}) \cdot(\vec{b}+\vec{c})=168\). Then the maximum value of \(|\vec{c}|^2\) is :

Let $A B C D$ be a tetrahedron such that the edges $\mathrm{AB}, \mathrm{AC}$ and AD are mutually perpendicular. Let the areas of the triangles $\mathrm{ABC}, \mathrm{ACD}$ and ADB be 5,6 and 7 square units respectively. Then the area (in square units) of the $\triangle \mathrm{BCD}$ is equal to :

Consider the lines $\mathrm{L}_1: \mathrm{x}-1=\mathrm{y}-2=\mathrm{z}$ and $\mathrm{L}_2: \mathrm{x}-2=\mathrm{y}=\mathrm{z}-1$. Let the feet of the perpendiculars from the point $\mathrm{P}(5,1,-3)$ on the lines $\mathrm{L}_1$ and $\mathrm{L}_2$ be $Q$ and $R$ respectively. If the area of the triangle PQR is A , then $4 \mathrm{~A}^2$ is equal to :

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 $(0,0),\left(\lambda_1, \lambda_2\right)$ and $\left(\lambda_2, \lambda_1\right)$ is