Vectors & 3D Geometry PYQ — Page 3

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{\mathrm{z}-2}{3}$. The coordinates of A and B are $(1,6,3)$ and $(4,9, \alpha)$ respectively and C is at a distance of 10 units from B. The area (in sq. units) of $\triangle \mathrm{ABC}$ is :

If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\dfrac{x}{1} = \dfrac{y-1}{1} = \dfrac{z-2}{2}$ is $4$, then the sum of all possible values of $a$ is equal to :

Let a line $L$ be perpendicular to both the lines $L_1: \dfrac{x+1}{3} = \dfrac{y+3}{5} = \dfrac{z+5}{7}$ and $L_2: \dfrac{x-2}{1} = \dfrac{y-4}{4} = \dfrac{z-6}{7}$. If $\theta$ is the acute angle between the lines $L$ and $L_3: \dfrac{x - \dfrac{8}{7}}{2} = \dfrac{y - \dfrac{4}{7}}{1} = \dfrac{z}{2}$, then $\tan\theta$ is equal to:

If the image of the point $\mathrm{P}(a, 2, a)$ in the line $\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}$ is Q and the image of Q in the line $\frac{x-2 b}{2}=\frac{y-a}{1}=\frac{z+2 b}{-5}$ is P, then $a+b$ is equal to $\_\_\_\_$.

Let the point A be the foot of perpendicular drawn from the point $P(a, b, 0)$ on the line $\dfrac{x-1}{2} = \dfrac{y-2}{1} = \dfrac{z-\alpha}{3}$. If the midpoint of the line segment PA is $\left(0, \dfrac{3}{4}, \dfrac{-1}{4}\right)$, then the value of $a^2 + b^2 + \alpha^2$ is equal to:

Let $\vec{a}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $\vec{c}=\vec{a} \times \vec{b}$. Let $\vec{d}$ be a vector such that $|\vec{d}-\vec{a}|=\sqrt{11},|\vec{c} \times \vec{d}|=3$ and the angle between $\vec{c}$ and $\vec{d}$ is $\frac{\pi}{4}$. Then $\vec{a} \cdot \vec{d}$ is equal to

Let the line $L$ pass through the point $(-3,5,2)$ and make equal angles with the positive coordinate axes. If the distance of L from the point $(-2, \mathrm{r}, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of r is :

Let $(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5,4,2)$ on the line $\overrightarrow{\mathrm{r}}=(-\hat{i}+3 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})$. Then the length of the projection of the vector $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ on the vector $6 \hat{i}+2 \hat{j}+3 \hat{k}$ is :

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

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

Let $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\mathrm{L}_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$ ?

The square of the distance of the point $\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ from the line $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ in the direction of the vector $\hat{i}+4 \hat{j}+7 \hat{k}$ is :

Let the values of p , for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{\mathrm{r}}=(\mathrm{p} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$ is $\frac{1}{\sqrt{6}}$, be $\mathrm{a}, \mathrm{b}$, $(a \lt b)$. Then the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ 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 $\vec{a}=2 \hat{i}-3 \hat{j}+k, \vec{b}=3 \hat{i}+2 \hat{j}+5 k$ and a vector $\vec{c}$ be such that $(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=-18 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \mathrm{k}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3$. If $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{d}}$, then $|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}|$ is equal to :

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 $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}, \vec{c}=\lambda \hat{j}+\mu \hat{k}$ and $\hat{d}$ be a unit vector such that $\overrightarrow{\mathrm{a}} \times \hat{\mathrm{d}}=\overrightarrow{\mathrm{b}} \times \hat{\mathrm{d}}$ and $\overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{d}}=1$, If $\vec{c}$ is perpendicular to $\vec{a}$, then $|3 \lambda \hat{d}+\mu \overrightarrow{\mathrm{c}}|^2$ is equal to _______ .

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 $A B C D$ 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 :

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 line passing through the points $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ intersect the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4,-5,1)$ is

Let the line L pass through $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$ . Then, which of the following points lies on the line L ?

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 $\overrightarrow{\mathrm{a}}=3 \hat{i}-\hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=\overrightarrow{\mathrm{a}} \times(\hat{i}-2 \hat{k})$ and $\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \hat{k}$. Then the projection of $\overrightarrow{\mathrm{c}}-2 \hat{j}$ on $\vec{a}$ is :