Chapter 26
71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers · 116 exercises
Problem 1
If the length of the perpendicular from the point \((\beta, 0, \beta)(\beta\) \(\neq 0\) ) to the line, \(\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}\) is \(\sqrt{\frac{3}{2}}\), then \(\beta\) is equal to: [April 10, 2019 (I)] (a) 1 (b) 2 (c) \(-1\) (d) \(-2\)
6 step solution
Problem 2
The vertices \(\mathrm{B}\) and \(\mathrm{C}\) of a "ABC lie on the line, \(\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}\) such that \(\mathrm{BC}=5\) units. Then the area (in sq. units) of this triangle, given that the point \(\mathrm{A}(1,-1,2)\), is: [April 09, 2019 (II)] (a) \(5 \sqrt{17}\) (b) \(2 \sqrt{34}\) (c) 6 (d) \(\sqrt{34}\)
6 step solution
Problem 3
If a point \(\mathrm{R}(4, y, z)\) lies on the line segment joining the points \(\mathrm{P}(2,-3,4)\) and \(\mathrm{Q}(8,0,10)\), then distance of \(\mathrm{R}\) from the origin is : [April 08, 2019 (II)] (a) \(2 \sqrt{14}\) (b) \(2 \sqrt{21}\) (c) 6 (d) \(\sqrt{53}\)
5 step solution
Problem 4
A tetrahedron has vertices \(\mathrm{P}(1,2,1), \mathrm{Q}(2,1,3), \mathrm{R}(-1,1,2)\) and \(\mathrm{O}(0,0,0) .\) The angle between the faces \(\mathrm{OPQ}\) and PQR is: [Jan. 12, 2019 (I)] (a) \(\cos ^{-1}\left(\frac{17}{31}\right)\) (b) \(\cos ^{-1}\left(\frac{19}{35}\right)\) (c) \(\cos ^{-1}\left(\frac{9}{35}\right)\) (d) \(\cos ^{-1}\left(\frac{7}{31}\right)\)
8 step solution
Problem 5
The length of the projection of the line segment joining the points \((5,-1,4)\) and \((4,-1,3)\) on the plane, \(x+y+z=7\) is:(a) \(\frac{2}{3}\) (b) \(\frac{1}{3}\) (c) \(\sqrt{\frac{2}{3}}\) (d) \(\frac{2}{\sqrt{3}}\)
5 step solution
Problem 6
An angle between the lines whose direction cosines are given by the equations, \(l+3 m+5 n=0\) and \(5 l m-2 m n+6 n l\) \(=0\), is [Online April 15, 2018] (a) \(\cos ^{-1}\left(\frac{1}{8}\right)\) (b) \(\cos ^{-1}\left(\frac{1}{6}\right)\) (c) \(\cos ^{-1}\left(\frac{1}{3}\right)\) (d) \(\cos ^{-1}\left(\frac{1}{4}\right)\)
7 step solution
Problem 7
\(\mathrm{ABC}\) is triangle in a plane with vertices \(\mathrm{A}(2,3,5), \mathrm{B}(-1,3,\), 2) and \(\mathrm{C}(\lambda, 5, \mu)\). If the median through A is equally inclined to the coordinate axes, then the value of \(\left(\lambda^{3}+\mu^{3}+5\right)\) is : [Online April 10, 2016] (a) 1130 (b) 1348 (c) 1077 (d) 676
6 step solution
Problem 8
The angle between the lines whose direction cosines satisfy the equations \(l+m+n=0\) and \(l^{2}+m^{2}+n^{2}\) is [2014] (a) \(\frac{\pi}{6}\) (b) \(\frac{\pi}{2}\) (c) \(\frac{\pi}{3}\) (d) \(\frac{\pi}{4}\)
6 step solution
Problem 9
Let \(\mathrm{A}(2,3,5), \mathrm{B}(-1,3,2)\) and \(\mathrm{C}(\lambda, 5, \mu)\) be the vertices of a \(\Delta \mathrm{ABC}\). If the median through \(\mathrm{A}\) is equally inclined to the coordinate axes, then: \(\quad\) [Online April 11, 2014] (a) \(5 \lambda-8 \mu=0\) (b) \(8 \lambda-5 \mu=0\) (c) \(10 \lambda-7 \mu=0\) (d) \(7 \lambda-10 \mu=0\)
5 step solution
Problem 10
A line in the 3-dimensional space makes an angle \(\theta\) \(\left(0<\theta \leq \frac{\pi}{2}\right)\) with both the \(x\) and y axes. Then the set of all values of \(\theta\) is the interval: \(\quad\) [Online April 9, 2014] (a) \(\left(0, \frac{\pi}{4}\right]\) (b) \(\left[\frac{\pi}{6}, \frac{\pi}{3}\right]\)(c) \(\left[\frac{\pi}{4}, \frac{\pi}{2}\right]\) (d) \(\left(\frac{\pi}{3}, \frac{\pi}{2}\right]\)
6 step solution
Problem 12
If the projections of a line segment on the \(x, y\) and \(z\)-axes in 3 -dimensional space are 2,3 and 6 respectively, then the length of the line segment is: [Online April \(23, \mathbf{2 0 1 3}]\) (a) 12 (b) 7 (c) 9 (d) 6
4 step solution
Problem 13
The acute angle between two lines such that the direction cosines \(l, m, n\), of each of them satisfy the equations \(l+m+n=0\) and \(l^{2}+m^{2}-n^{2}=0\) is : [Online April 22, 2013] (a) \(15^{\circ}\) (b) \(30^{\circ}\) (c) \(60^{\circ}\) (d) \(45^{\circ}\)
5 step solution
Problem 14
A line \(\mathrm{AB}\) in three-dimensional space makes angles \(45^{\circ}\) and \(120^{\circ}\) with the positive \(x\)-axis and the positive \(y\)-axis respectively. If AB makes an acute angle \(\theta\) with the positive z-axis, then \(\theta\) equals [2010] (a) \(45^{\circ}\) (b) \(60^{\circ}\) (c) \(75^{\circ}\) (d) \(30^{\circ}\)
5 step solution
Problem 15
The projections of a vector on the three coordinate axis are \(6,-3,2\) respectively. The direction cosines of the vector are [2009] (a) \(\frac{6}{5}, \frac{-3}{5}, \frac{2}{5}\) (b) \(\frac{6}{7}, \frac{-3}{7}, \frac{2}{7}\) (c) \(\frac{-6}{7}, \frac{-3}{7}, \frac{2}{7}\) (d) \(6,-3,2\)
4 step solution
Problem 17
A line makes the same angle \(\theta\), with each of the \(x\) and \(z\) axis. If the angle \(\beta\), which it makes with \(y\)-axis, is such that \(\sin ^{2} \beta=3 \sin ^{2} \theta\), then \(\cos ^{2} \theta\) equals \(\quad\) [2004] (a) \(\frac{2}{5}\) (b) \(\frac{1}{5}\) (c) \(\frac{3}{5}\) (d) \(\frac{2}{3}\)
7 step solution
Problem 18
A plane \(\mathrm{P}\) meets the coordinate axes at \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) respectively. The centroid of \(\Delta \mathrm{ABC}\) is given to be \((1,1,2)\). Then the equation of the line through this centroid and perpendicular to the plane \(\mathrm{P}\) is: [Sep. 06, 2020 (II)] (a) \(\frac{x-1}{2}=\frac{y-1}{1}=\frac{z-2}{1}\) (b) \(\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-2}{2}\) (c) \(\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1}\) (d) \(\frac{x-1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\)
6 step solution
Problem 20
The lines \(\vec{r}=(\hat{i}-\hat{j})+l(2 \hat{i}+\hat{k})\) and \(\vec{r}=(2 \hat{i}-\hat{j})+m(\hat{i}+\hat{j}-\hat{k}) \quad\) [Sep. 03, 2020 (I)] (a) do not intersect for any values of \(l\) and \(m\) (b) intersect for all values of \(l\) and \(m\) (c) intersect when \(l=2\) and \(m=\frac{1}{2}\) (d) intersect when \(l=1\) and \(m=2\)
5 step solution
Problem 21
The shortest distance between the lines \(\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}\) and \(\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}\) is: [Jan. 08, 2020 (I)] (a) \(2 \sqrt{30}\) (b) \(\frac{7}{2} \sqrt{30}\) (c) \(3 \sqrt{30}\) (d) 3
4 step solution
Problem 23
A perpendicular is drawn from a point on the line \(\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}\) to the plane \(x+y+z=3\) such that the foot of the perpendicular \(\mathrm{Q}\) also lies on the plane \(x-y+z=3\). Then the co-ordinates of Q are : [April 10, 2019 (II)](a) \((1,0,2)\) (b) \((2,0,1)\) (c) \((-1,0,4)\) (d) \((4,0,-1)\)
6 step solution
Problem 24
The length of the perpendicular from the point \((2,-1,4)\) on the straight line, \(\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}\) is [April \(\mathbf{0 8}, \mathbf{2 0 1 9}\) (I)] (a) greater than 3 but less than 4 (b) less than 2 (c) greater than 2 but less than 3 (d) greater than 4
7 step solution
Problem 25
Two lines \(\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}\) and \(\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}\) intersect at the point \(\mathrm{R}\). The reflection of \(\mathrm{R}\) in the \(x y\) - plane has coordinates : \(\quad\) [Jan.11, 2019 (II)] (a) \((2,-4,-7)\) (b) \((2,4,7)\) (c) \((2,-4,7)\) (d) \((-2,4,7)\)
7 step solution
Problem 27
If the angle between the lines, \(\frac{x}{2}=\frac{y}{2}=\frac{z}{1}\) and \(\frac{5-x}{-2}=\frac{7 y-14}{P}=\frac{z-3}{4}\) is \(\cos ^{-1}\left(\frac{2}{3}\right)\), then \(\mathrm{P}\) is equal to \([\) Online April \(16, \mathbf{2 0 1 8}]\) (a) \(-\frac{7}{4}\) (b) \(\frac{2}{7}\) (c) \(-\frac{4}{7}\) (d) \(\frac{7}{2}\)
7 step solution
Problem 28
The number of distinct real values of \(\lambda\) for which the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda^{2}}\) and \(\quad \frac{x-3}{1}=\frac{y-2}{\lambda^{2}}=\frac{z-1}{2} \quad\) are coplanar is : \(\quad\) [Online April 10, 2016] (a) 2 (b) 4 (c) 3 (d) 1
5 step solution
Problem 29
The shortest distance between the lines \(\frac{x}{2}=\frac{y}{2}=\frac{z}{1}\) and \(\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\) lies in the interval : [Online April 9, 2016] (a) \((3,4]\) (b) \((2,3]\) (c) \([1,2)\) (d) \([0,1)\)
7 step solution
Problem 30
Equation of the line of the shortest distance between the lines \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{1}\) and \(\frac{\mathrm{x}-1}{0}=\frac{\mathrm{y}+1}{-2}=\frac{\mathrm{z}}{1}\) is: [Online April 19, 2014] (a) \(\frac{\mathrm{x}}{\mathrm{l}}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{-2}\) (b) \(\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z}{-2}\) (c) \(\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z}{1}\) (d) \(\frac{\mathrm{x}}{-2}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}}{2}\)
4 step solution
Problem 31
If the lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) and \(\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}\) are coplanar, then \(k\) can have [2013] (a) any value (b) exactly one value (c) exactly two values (d) exactly three values
8 step solution
Problem 32
If two lines \(L_{1}\) and \(L_{2}\) in space, are defined by \(L_{1}=\\{x=\sqrt{\lambda} y+(\sqrt{\lambda}-1), z=(\sqrt{\lambda}-1) y+\sqrt{\lambda}\\}\) and \(L_{2}=\\{x=\sqrt{\mu} y+(1-\sqrt{\mu}), z=(1-\sqrt{\mu}) y+\sqrt{\mu}\\}\) then \(L_{1}\) is perpendicular to \(L_{2}\), for all non-negative reals \(\lambda\) and \(\mu\), such that : \(\quad\) [Online April 23, 2013] (a) \(\sqrt{\lambda}+\sqrt{\mu}=1\) (b) \(\lambda \neq \mu\) (c) \(\lambda+\mu=0\) (d) \(\lambda=\mu\)
5 step solution
Problem 33
If the lines \(\frac{x+1}{2}=\frac{y-1}{1}=\frac{z+1}{3}\) and \(\frac{x+2}{2}=\frac{y-k}{3}=\frac{z}{4}\) are coplanar, then the value of \(k\) is : [Online April 9, 2013] (a) \(\frac{11}{2}\) (b) \(-\frac{11}{2}\) (c) \(\frac{9}{2}\) (d) \(-\frac{9}{2}\)
5 step solution
Problem 34
If the line \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\) intersect, then \(k\) is equal to: (a) \(-1\) (b) \(\frac{2}{9}\) (c) \(\frac{9}{2}\) (d) 0
6 step solution
Problem 35
The distance of the point \(-\hat{i}+2 \hat{j}+6 \hat{k}\) from the straight line that passes through the point \(2 \hat{i}+3 \hat{j}-4 \hat{k}\) and is parallel to the vector \(6 \hat{i}+3 \hat{j}-4 \hat{k}\) is [Online May 26, 2012]
6 step solution
Problem 37
The coordinates of the foot of perpendicular from the point \((1,0,0)\) to the line \(\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}\) are \(\quad\) [Online May 12, 2012] (a) \((2,-3,8)\) (b) \((1,-1,-10)\) (c) \((5,-8,-4)\) (d) \((3,-4,-2)\)
7 step solution
Problem 38
The length of the perpendicular drawn from the point \((3,-1,11)\) to the line \(\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) is: \(\quad\) [2011RS] (a) \(\sqrt{29}\) (b) \(\sqrt{33}\) (c) \(\sqrt{53}\) (d) \(\sqrt{66}\)
8 step solution
Problem 39
Statement-1: The point \(A(1,0,7)\) is the mirror image of the point \(B(1,6,3)\) in the line: \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}\) Statement-2: The line \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}\) bisects the line segment joining \(A(1,0,7)\) and \(B(1,6,3)\). (a) Statement- 1 is true, Statement- 2 is true; Statement- 2 is not a correct explanation for Statement- 1 . (b) Statement- 1 is true, Statement- 2 is false. (c) Statement- 1 is false, Statement- 2 is true. (d) Statement- 1 is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement-1.
3 step solution
Problem 41
If the straight lines \(\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}\) intersect at a point, then the integer \(\mathrm{k}\) is equal to [2008] (a) \(-5\) (b) 5 (c) 2 (d) \(-2\)
4 step solution
Problem 42
If non zero numbers \(a, b, c\) are in H.P., then the straight line \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\) always passes through a fixed point. That point is [2005] (a) \((-1,2)\) (b) \((-1,-2)\) (c) \((1,-2)\) (d) \(\left(1,-\frac{1}{2}\right)\)
5 step solution
Problem 44
If the straight lines [2004] \(x=1+s, y=-3-\lambda s, z=1+\lambda s\) and \(x=\frac{t}{2}, y=1+t, z=2-t\), with parameters \(\mathrm{s}\) and \(\mathrm{t}\) respectively, are co-planar, then \(\lambda\) equals. (a) 0 (b) \(-1\) (c) \(-\frac{1}{2}\) (d) \(-2\)
5 step solution
Problem 45
A line with direction cosines proportional to \(2,1,2\) meets each of the lines \(x=y+a=z\) and \(x+a=2 y=2 z\). The co-ordinates of each of the points of intersection are given by \([2004]\) (a) \((2 a, 3 a, 3 a),(2 a, a, a)\) (b) \((3 a, 2 a, 3 a),(a, a, a)\) (c) \((3 a, 2 a, 3 a),(a, a, 2 a)\) (d) \((3 a, 3 a, 3 a),(a, a, a)\)
5 step solution
Problem 46
The lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) and \(\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}\) are coplanar if (a) \(\mathrm{k}=3\) or \(-2\) (b) \(\mathrm{k}=0\) or \(-1\) (c) \(\mathrm{k}=1\) or \(-1\) (d) \(\mathrm{k}=0\) or \(-3\).
6 step solution
Problem 48
The shortest distance between the lines \(\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}\) and \(x+y+z+1=0,2 x-y+z+3=0\) is: \([\operatorname{Sep} .06,2020(I)]\) (a) 1 (b) \(\frac{1}{\sqrt{3}}\) (c) \(\frac{1}{\sqrt{2}}\) (d) \(\frac{1}{2}\)
7 step solution
Problem 49
The shortest distance between the lines \(\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}\) and \(x+y+z+1=0,2 x-y+z+3=0\) is: \([\operatorname{Sep} .06,2020(I)]\) (a) 1 (b) \(\frac{1}{\sqrt{3}}\) (c) \(\frac{1}{\sqrt{2}}\) (d) \(\frac{1}{2}\)
6 step solution
Problem 50
If the equation of a plane \(P\), passing through the intersection of the planes, \(x+4 y-z+7=0\) and \(3 x+y+5 z\) \(=8\) is \(a x+b y+6 z=15\) for some \(a, b \in \mathbf{R}\), then the distance of the point \((3,2,-1)\) from the plane \(P\) is
5 step solution
Problem 51
The distance of the point \((1,-2,3)\) from the plane \(x-y+z=5\) measured parallel to the line \(\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}\) is: \(\quad\) [NA Sep.04, 2020 (II)] (a) \(\frac{7}{5}\) (b) 1 (c) \(\frac{1}{7}\) (d) 7
4 step solution
Problem 52
The foot of the perpendicular drawn from the point \((4,2,3)\) to the line joining the points \((1,-2,3)\) and \((1,1,0)\) lies on the plane: [Sep. 03, 2020 (I)] (a) \(2 x+y-z=1\) (b) \(x-y-2 z=1\) (c) \(x-2 y+z=1\) (d) \(x+2 y-z=1\)
5 step solution
Problem 53
The plane which bisects the line joining the points \((4,-2,3)\) and \((2,4,-1)\) at right angles also passes through the point: \([\) Sep. \(03,2020(\mathrm{II})]\) (a) \((4,0,1)\) (b) \((0,-1,1)\) (c) \((4,0,-1)\) (d) \((0,1,-1)\)
4 step solution
Problem 54
Let a plane \(P\) contain two lines \(\vec{r}=\hat{i}+\lambda(\hat{i}+\hat{j}), \lambda \in \mathbf{R}\) and \(\vec{r}=-\hat{j}+\mu(\hat{j}-\hat{k}), \mu \in \mathbf{R}\) If \(Q(\alpha, \beta, \gamma)\) is the foot of the perpendicular drawn from the point \(M(1,0,1)\) to \(P\), then \(3(\alpha+\beta+\gamma)\) equals
5 step solution
Problem 56
A plane passing through the point \((3,1,1)\) contains two lines whose direction ratios are \(1,-2,2\) and \(2,3,-1\) respectively. If this plane also passes through the point \((\alpha,-3,5)\), then \(\alpha\) is equal to : \(\quad\) [Sep. 02, 2020 (II)] (a) 5 (b) \(-10\) (c) 10 (d) \(-5\)
4 step solution
Problem 57
If for some \(\alpha\) and \(\beta\) in \(\mathbf{R}\), the intersection of the following three planes \(x+4 y-2 z=1\) \(x+7 y-5 z=\beta\) \(x+5 y+\alpha z=5\) is a line in \(\mathrm{R}^{3}\), then \(\alpha+\beta\) is equal to: \(\quad\) [Jan. \(9, \mathbf{2 0 2 0}\) (I)] (a) 0 (b) 10 (c) 2 (d) \(-10\)
7 step solution
Problem 58
If the distance between the plane, \(23 x-10 y-2 z+48=0\) and the plane containing the lines \(\frac{x+1}{2}=\frac{y-3}{4}=\frac{z+1}{3}\) and \(\frac{x+3}{2}=\frac{y+2}{6}=\frac{z-1}{\lambda}(\lambda \in \mathbf{R})\) is equal to \(\frac{k}{\sqrt{633}}\), then \(k\) is equal to [NA Jan. 9, 2020 (II)]
5 step solution
Problem 60
Let \(P\) be a plane passing through the points \((2,1,0)\), \((4,1,1)\) and \((5,0,1)\) and \(R\) be any point \((2,1,6)\). Then the image of \(R\) in the plane \(P\) is: \(\quad\) [Jan. 7, 2020 (I)] (a) \((6,5,2)\) (b) \((6,5,-2)\) (c) \((4,3,2)\) (d) \((3,4,-2)\)
6 step solution
Problem 61
If the line \(\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}\) intersects the plane \(2 x+3 y-z+13=0\) at a point \(P\) and the plane \(3 x+y+4 z=\) 16 at a point \(Q\), then \(P Q\) is equal to: [April 12, 2019 (I)] (a) 14 (b) \(\sqrt{14}\) (c) \(2 \sqrt{7}\) (d) \(2 \sqrt{14}\)
7 step solution