JEE 3D Geometry PYQs with Solutions PDF, Download Now

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

For a triangle ABC, let $$\overrightarrow{p} = \overrightarrow{BC}, \overrightarrow{q}= \overrightarrow{CA}$$ and $$\overrightarrow{r} = \overrightarrow{BA}$$. If $$|\overrightarrow{p}| = 2\sqrt{3}, |\overrightarrow{q}|=2$$ and $$\cos\theta = \frac{1}{\sqrt{3}}$$ where $$\theta$$ is the angle between $$\overrightarrow{p}$$ and $$\overrightarrow{q}$$, then $$|\overrightarrow{p} \times \left(\overrightarrow{q}-\overrightarrow{3r}\right)|^2 +3|\overrightarrow{r}|^2$$ 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, r, 1) is $$\sqrt{\frac{14}{3}}$$, then the sum of all possible values of r is:

Let L be the line $$\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$$ and let S be the set of all points (a, b, c) on L, whose distance from the line $$\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$$a long the line L is 7. Then $$\sum_{(a,b,c)\in S} (a+b+c) $$ is equal to :

Let a vector $$\vec{a} = \sqrt{2}\,\hat{i} - \hat{j} + \lambda \hat{k}, \quad \lambda > 0,$$ make an obtuse angle with the vector $$\vec{b} = -\lambda^{2}\hat{i} + 4\sqrt{2}\,\hat{j} + 4\sqrt{2}\,\hat{k}$$ and an angle $$\theta, \dfrac{\pi}{6} < \theta < \dfrac{\pi}{2}$$, with the positive z-axis. If the set of all possible values of $$\lambda$$ is $$( \alpha, \beta) - \{\gamma\}$$, then $$\alpha + \beta + \gamma$$ is equal to __________.

The vertices B and C of a triangle ABC lie on the line $$\frac{x}{1}=\frac{1-y}{-2}=\frac{z-2}{3}$$ 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 $$\triangle$$ABC is :

Let the direction cosines of two lines satisfy the equations : 4l + m - n =0 and 2mn + l0nl +3lm= 0. Then the cosine of the acute angle between these lines is :

If the image of the point $$P(1 , 2, a)$$ in the line $$ \frac{x-6}{3} = \frac{y - 7}{2} = \frac{7 -z}{2}$$ is $$Q(5, b, c)$$, then $$ a^{2}+b^{2}+c^{2}$$ is equal to

Let $$ P(\alpha,\beta, \gamma)$$ be the point on the line $$\frac{x-1}{2}=\frac{y+1}{-3}=z$$ at a distance $$4\sqrt{14}$$ from the point (1, -1, 0) and nearer to the origin. Then the shortest di stance, between the Lines $$\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$$ and $$\frac{x+5}{2}= \frac{y-10}{1}=\frac{z-3}{1}$$, is equal to

Let the lines $$L_{1}:\overrightarrow{r}=\widehat{i}+2\widehat{j}+3\widehat{k}+\lambda(2\widehat{i}+3\widehat{j}+4\widehat{k}),\lambda \in \mathbb{R}$$ and $$L_{2}:\overrightarrow{r}=(4\widehat{i}+\widehat{j})+\mu (5\widehat{i}+2\widehat{j}+\widehat{k}),\mu \in \mathbb{R}$$, interest at the point R. Let P and Q be the points lying on lines $$L_{1}$$ and $$L_{2}$$ respectively, such that $$|\overrightarrow{PR}|=\sqrt{29}$$ and $$|\overrightarrow{PQ}|=\sqrt{\frac{47}{3}}$$. If the point P lies in the first octant, then $$27(QR)^{2}$$ is equal to

Let a line L passing through the point P (1, 1, 1) be perpendicular to the lines $$\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$$ and $$\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$$. Let the line L intersect the yz-plane at the point Q. Another line parallel to L and passing through the point S (1, 0, - 1) intersects the yz-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to ___.

The sum of all values of $$\alpha$$, for which the sho1test distance between the lines
$$\frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha}$$ and $$\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha}$$ is $$\sqrt{2}$$, is

If the distances of the point (1 , 2, a) from the line $$\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$$ along the lines $$L_{1}:\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$$ and $$L_{2}:\frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$$ are equal, then a + b + c is equal to

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

Let P be a point in the plane of the vectors $$ \overrightarrow{AB}=3\widehat{i} + \widehat{j}-\widehat{k} \text{ and }\overrightarrow{AC}=\widehat{i}-\widehat{j}+3\widehat{k}$$ such that P is equidistant from the Lines AB and AC. If $$ \mid \overrightarrow{AP} \mid=\frac{\sqrt{5}}{2} $$ then the area of the triangle ABP is:

If the distance of the point $$P(43, \alpha, \beta), \beta<0,$$ from the line $$\overrightarrow{r} = 4\widehat{i}-\widehat{k}+\mu(2\widehat{i}+3\widehat{k}), \mu \in \mathbb{R}$$ along a line with direction ratios 3, -1, 0 is $$13\sqrt{10},$$ then $$ \alpha ^{2}+ \beta^{2}$$ is equal to______

Let a line pass through two distinct points P(−2,−1, 3) and , and be parallel to the vector $$3\widehat{i}+2\widehat{j}+2\widehat{k}$$. If the distance of the point Q from the point R(1, 3, 3) is 5 , then the square of the area of △PQR is equal to :

The perpendicular distance, of the line $$\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}$$ from the point P(2,−10, 1), is :

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

Let $$L_1:\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$$ and $$L_2:\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}$$, $$\alpha \in R$$, 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 $$26\alpha(PB)^{2}$$ is ______

The distance of the line $$\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$$ from the point (1, 4, 0) along the line $$\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$$ is :

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

Let the point A divide the line segment joining the points P(−1,−1, 2) and Q(5, 5, 10) internally in the ratio $$r : 1 (r > 0)$$. If O is the origin and $$(\overrightarrow{OQ}.\overrightarrow{OA})-\frac{1}{5}|\overrightarrow{OP}.\overrightarrow{OA}|^{2}=10$$. then the value of r is :

Let in a $$\triangle ABC$$, the length of the side $$AC$$ 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 $$\triangle ABC$$ 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  $$P$$  be the image of the point  $$Q(7,-2,5)$$ in the line  $$L:\;\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4},$$ and  $$R(5,p,q)$$  be a point on $$L.$$ Then the square of the area of  $$\triangle PQR$$ is  $$\underline{\hspace{2cm}}.$$

If the image of the point (4,4,3)in the line $$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$$ is $$(\alpha ,\beta ,\gamma)$$, then $$\alpha +\beta +\gamma$$ is equal to

Let $$A(x,y,Z)$$ be a point in $$xy-plain$$,which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B = (1, 4, −1) and C = (2, 0, −2). Then among the statements $$(SI): \triangle ABC$$ is an isosceles right angled triangle, and (SI):the area of $$\triangle ABC$$ is $$\frac{9\sqrt{2}}{2}$$,

Let $$L_{1}: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$$ and $$L_{2}: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$$ be two lines. Let $$L_{3}$$ be a line passing through the point $$(\alpha ,\beta ,\gamma)$$ and be perpendicular to both $$L_{1}$$ and $$L_{2}$$. If $$L_{3}$$ intersects $$L_{1}$$, then $$|5\alpha -11\beta -8\gamma|$$ equals:

If the system of equations $$(\lambda-1)x+(\lambda-4)y+\lambda z=5 \\\lambda x+(\lambda-1)y+(\lambda-4)z=7 \\ (\lambda+1)x+(\lambda+2)y-(\lambda+2)z=9$$ has infinitely many solutions, then $$\lambda^{2}+\lambda$$ is equal to

Let a straight line L pass through the point P(2,-1,3) and be perpendicular to the lines $$\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-2}$$ and $$\frac{x-3}{1}=\frac{y-2}{3}=\frac{z+2}{4}.$$ If the line L intersects the yz-plane at the point Q , then the distance between the points P and Q is :

Let the point, on the line passing through the points P(1, −2, 3) and Q(5, −4, 7), farther from the origin and at distance of 9 units from the point P, be $$(\alpha, \beta, \gamma)$$. Then $$\alpha^2 + \beta^2 + \gamma^2$$ is equal to:

If the shortest distance between the lines $$\frac{x+2}{2} = \frac{y+3}{3} = \frac{z-5}{4}$$ and $$\frac{x-3}{1} = \frac{y-2}{-3} = \frac{z+4}{2}$$ is $$\frac{38}{3\sqrt{5}}k$$, and $$\int_0^k [x^2]dx = \alpha - \sqrt{\alpha}$$, where [x] denotes the greatest integer function, then $$6\alpha^3$$ is equal to ______.

If the components of $$\overrightarrow{a} = \alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$$ along and perpendicular to $$\overrightarrow{b}= 3\hat{i}+\hat{j}-\hat{k}$$ 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 equals to :

The square of the distance of the point $$(\frac{15}{7},\frac{32}{7},7)$$ 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 :

If the system of equations $$\\2x + 3y - z = 5\\ x + \alpha y + 3z = -4\\ 3x - y + \beta z = 7\\$$ has infinitely many solutions, then $$13\alpha\beta$$ is equal to:

If the shortest distance between the lines $$\frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1}$$ and $$\frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1}$$ is $$1$$, then the sum of all possible values of $$\lambda$$ is:

Let the line of the shortest distance between the lines $$L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$$ and $$L_2: \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$$ intersect $$L_1$$ and $$L_2$$ at $$P$$ and $$Q$$ respectively. If $$(\alpha, \beta, \gamma)$$ is the midpoint of the line segment $$PQ$$, then $$2(\alpha + \beta + \gamma)$$ is equal to:

Let the system of equations $$x + 2y + 3z = 5$$, $$2x + 3y + z = 9$$, $$4x + 3y + \lambda z = \mu$$ have infinite number of solutions. Then $$\lambda + 2\mu$$ is equal to:

Consider a $$\triangle ABC$$ where $$A(1, 3, 2)$$, $$B(-2, 8, 0)$$ and $$C(3, 6, 7)$$. If the angle bisector of $$\angle BAC$$ meets the line BC at D, then the length of the projection of the vector $$\vec{AD}$$ on the vector $$\vec{AC}$$ is:

If the mirror image of the point $$P(3, 4, 9)$$ in the line $$\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$$ is $$(\alpha, \beta, \gamma)$$, then $$14(\alpha + \beta + \gamma)$$ is:

Let P and Q be the points on the line $$\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$$ which are at a distance of 6 units from the point $$R(1, 2, 3)$$. If the centroid of the triangle PQR is $$(\alpha, \beta, \gamma)$$, then $$\alpha^2 + \beta^2 + \gamma^2$$ is:

The distance, of the point $$(7, -2, 11)$$ from the line $$\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$$ along the line $$\frac{x-5}{2} = \frac{y-1}{-3} = \frac{z-5}{6}$$, is :

If the shortest distance between the lines $$\frac{x-4}{1} = \frac{y+1}{2} = \frac{z}{-3}$$ and $$\frac{x-\lambda}{2} = \frac{y+1}{4} = \frac{z-2}{-5}$$ is $$\frac{6}{\sqrt{5}}$$, then the sum of all possible values of $$\lambda$$ is :

The position vectors of the vertices A, B and C of a triangle are $$2\hat{i} - 3\hat{j} + 3\hat{k}$$, $$2\hat{i} + 2\hat{j} + 3\hat{k}$$ and $$-\hat{i} + \hat{j} + 3\hat{k}$$ respectively. Let $$l$$ denotes the length of the angle bisector AD of $$\angle BAC$$ where D is on the line segment BC, then $$2l^2$$ equals :

Let the image of the point $$(1, 0, 7)$$ in the line $$\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$$ be the point $$(\alpha, \beta, \gamma)$$. Then which one of the following points lies on the line passing through $$(\alpha, \beta, \gamma)$$ and making angles $$\frac{2\pi}{3}$$ and $$\frac{3\pi}{4}$$ with y-axis and z-axis respectively and an acute angle with x-axis?

The lines $$\frac{x-2}{2} = \frac{y}{-2} = \frac{z-7}{16}$$ and $$\frac{x+3}{4} = \frac{y+2}{3} = \frac{z+2}{1}$$ intersect at the point P. If the distance of P from the line $$\frac{x+1}{2} = \frac{y-1}{3} = \frac{z-1}{1}$$ is $$l$$, then $$14l^2$$ is equal to _____.

Let $$PQR$$ be a triangle with $$R(-1, 4, 2)$$. Suppose $$M(2, 1, 2)$$ is the mid point of $$PQ$$. The distance of the centroid of $$\Delta PQR$$ from the point of intersection of the line $$\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1}$$ and $$\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$$ is

A line with direction ratio $$2, 1, 2$$ meets the lines $$x = y + 2 = z$$ and $$x + 2 = 2y = 2z$$ respectively at the point $$P$$ and $$Q$$. If the length of the perpendicular from the point $$(1, 2, 12)$$ to the line $$PQ$$ is $$l$$, then $$l^2$$ is _______

Let $$O$$ be the origin, and $$M$$ and $$N$$ be the points on the lines $$\frac{x-5}{4} = \frac{y-4}{1} = \frac{z-5}{3}$$ and $$\frac{x+8}{12} = \frac{y+2}{5} = \frac{z+11}{9}$$ respectively such that $$MN$$ is the shortest distance between the given lines. Then $$\vec{OM} \cdot \vec{ON}$$ is equal to ______.