JEE Mains JEE Advance BITSAT
JEE Mains | PYQ

Year

All years 2026 2025 2024 2023 2022 2021

If the image of the point \( P(1, 0, 3) \) in the line joining the points \( A(4, 7, 1) \) and \( B(3, 5, 3) \) is Q\((\alpha, \beta, \gamma)\), then \( \alpha + \beta + \gamma \) 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 \( \vec{r}=(\hat{i}-\hat{j}+6\hat{k})+\mu(-b\hat{i}+a\hat{j}+5\hat{k}) \) is \( \frac{x-1}{-2}=\frac{y-4}{d}=\frac{z-c}{-4} \), then \( a+b+c+d \) is equal to:

Let \( \vec a=\hat i+2\hat j+\hat k \) and \( \vec b=2\hat i+7\hat j+3\hat k \). Let \( L_1:\ \vec r=(-\hat i+2\hat j+\hat k)+\lambda \vec a,\, \lambda\in \mathbb{R} \) and \( L_2:\ \vec r=(\hat j+\hat k)+\mu \vec b,\, \mu\in \mathbb{R} \) be two lines. If the line \( L_3 \) passes through the point of intersection of \( L_1 \) and \( L_2 \), and is parallel to \( \vec a+\vec b \), then \( L_3 \) passes through the point:

Let a line passing through the point \( (4,1,0) \) intersect the line \( L_1:\; \dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4} \) at the point \( A(\alpha,\beta,\gamma) \) and the line \( L_2:\; x-6=y=-z+4 \) at the point \( B(a,b,c) \).

Let the values of \( p \), for which the shortest distance between the lines \( \frac{x+1}{3}=\frac{y}{4}=\frac{z}{5} \) and \( \vec{r}= (p\hat{i}+2\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}+4\hat{k}) \) is \( \frac{1}{\sqrt{6}} \), be \( a, b \), \( (a<b)\).Then the length of the latus rectum of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is:

Let \( A(x, y, z) \) be a point in \( xy \)-plane, 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: \( (S1) \): \( \triangle ABC \) is an isosceles right angled triangle, and \( (S2) \): the area of \( \triangle ABC \) is \( \dfrac{9\sqrt{2}}{2} \).

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

Line \( L_{1} \) passes through the point \( (1,2,3) \) and is parallel to \( z \)-axis. Line \( L_{2} \) passes through the point \( (\lambda,5,6) \) and is parallel to \( y \)-axis. Let for \( \lambda = \lambda_{1}, \lambda_{2}, \lambda_{2} < \lambda_{1} \), the shortest distance between the two lines be \( 3 \). Then the square of the distance of the point \( (\lambda_{1}, \lambda_{2}, 7) \) from the line \( L_{1} \) is:

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

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

Let the values of \( \lambda \) for which the shortest distance between the lines \( \dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4} \) and \( \dfrac{x-\lambda}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5} \) is \( \dfrac{1}{\sqrt{6}} \) 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:

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 \( 2\vec{p} - \vec{q} + 2\vec{r} \). 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:

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

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

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

Let the vertices \( Q \) and \( R \) of the triangle \( PQR \) lie on the line \( \dfrac{x+3}{5} = \dfrac{y-1}{2} = \dfrac{z+4}{3} \), \( QR = 5 \) and the coordinates of the point \( P \) be \( (0, 2, 3) \). If the area of the triangle \( PQR \) is \( \dfrac{m}{n} \) then:

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

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:

Each of the angles \( \beta \) and \( \gamma \) that a given line makes with the positive \( y \)- and \( z \)-axes, respectively, is half of the angle that this line makes with the positive \( x \)-axes. Then the sum of all possible values of the angle \( \beta \) is:

Let \( A \) be the point of intersection of the lines \( L_1 : \, \frac{x-7}{1} = \frac{y-5}{0} = \frac{z-3}{-1} \) and \( L_2 : \, \frac{x-1}{3} = \frac{y+3}{4} = \frac{z+7}{5} \).Let \( B \) and \( C \) be the points on the lines \( L_1 \) and \( L_2 \) respectively such that \( AB = AC = \sqrt{15} \). Then the square of the area of the triangle \( ABC \) 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 a straight line \( L \) pass through the point \( P(2,-1,3) \) and be perpendicular to the lines \(\displaystyle \frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-2} \) and \(\displaystyle \frac{x-3}{1}=\frac{y-2}{3}=\frac{z+2}{4} \). If the line \( L \) intersects the \( yz \)-plane at the point \( Q \), then the distance between the point \( P \) and \( Q \) is:

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

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:

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

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

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

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. Then which of the following points lies on the line of the shortest distance between \( L_{1} \) and \( L_{2} \)?

The distance of the point \( (7, 10, 11) \) from the line \( \frac{x-4}{1} = \frac{y-4}{0} = \frac{z-2}{3} \) along the line \( \frac{x-9}{2} = \frac{y-13}{3} = \frac{z-17}{6} \) is:

Question 1 of 32Next →

JEE Main 2025 2nd April Evening Shift

If the image of the point \( P(1, 0, 3) \) in the line joining the points \( A(4, 7, 1) \) and \( B(3, 5, 3) \) is Q\((\alpha, \beta, \gamma)\), then \( \alpha + \beta + \gamma \) is equal to: