Jee Math Test Series Dimensional Geometry Quiz-1

If the length of the perpendicular from the point \((b, 0, b)(b \neq 0)\) to the line, \(\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}\) is \(\sqrt{\frac{3}{2}}\), then \(\mathrm{b}\) is equal to:

The vertices \(\mathrm{B}\) and \(\mathrm{C}\) of a \(\triangle \mathrm{ABC}\) 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:

If a point \(\mathrm{R}(4, y, z)\) lies on the line segment joining the points \(\mathrm{P}(2,-3\), 4) and \(Q(8,0,10)\), then distance of \(R\) from the origin is :

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

If \((a, b, c)\) is the image of the point \((1,2,-3)\) in the line, \(\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}\), then \(a+b+c\) is equals to:

The lines \(\vec{r}=(\hat{i}-\hat{j})+l(2 \hat{i}+\hat{k})\) and \(\vec{r}=(2 \hat{i}-\hat{j})+m(\hat{i}+\hat{j}-\hat{k})\)

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:

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 :

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 :

If for some \(\alpha \in \mathbf{R}\), the lines \(L_1: \frac{x+1}{2}=\frac{y-2}{-1}=\frac{z-1}{1}\) and \(L_2: \frac{x+2}{\alpha}=\frac{y+1}{5-\alpha}=\frac{z+1}{1}\) are coplanar, then the line \(L_2\) passes through the point :

The perpendicular distance from the origin to the plane containing the two lines, \(\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}\) and \(\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}\), is :

Let \(S\) be the set of all real values of \(\lambda\) such that a plane passing through the points \(\left(-\lambda^2, 1,1\right),\left(1,-\lambda^2, 1\right)\) and \(\left(1,1,-\lambda^2\right)\) also passes through the point- \((-1,-1,1)\). Then \(S\) is equal to :

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 \(P\) is equal to

The angle between the lines whose direction cosines satisfy the equations \(l+m+n=0\) and \(l^2+m^2+n^2\) is

Let \(A(2,3,5), B(-1,3,2)\) and \(C(\lambda, 5, \mu)\) be the vertices of a \(\triangle \mathrm{ABC}\). If the median through \(\mathrm{A}\) is equally inclined to the coordinate axes, then:

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:

Let \(A B C\) be a triangle with vertices at points \(A(2,3,5)\), \(B(-1,3,2)\) and \(C(\lambda, 5, \mu)\) in three dimensional space. If the median through \(\mathrm{A}\) is equally inclined with the axes, then \((\lambda, \mu)\) is equal to :

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 :

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 :

A line \(A B\) in three-dimensional space makes angles \(45^{\circ}\) and \(120^{\circ}\) with the positive \(\mathrm{x}\)-axis and the positive \(\mathrm{y}\)-axis respectively. If \(A B\) makes an acute angle \(\theta\) with the positive \(z\)-axis, then \(\theta\) equals

The projections of a vector on the three coordinate axis are 6 , \(-3,2\) respectively. The direction cosines of the vector are :

If a line makes an angle of \(\pi / 4\) with the positive directions of each of \(x\)-axis and \(y\)-axis, then the angle that the line makes with the positive direction of the \(z\)-axis is

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

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 \(\frac{x-3}{1}=\frac{y-2}{\lambda^2}=\frac{z-1}{2}\) are coplanar is :

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 :

Equation of the line of the shortest distance between the lines \(\frac{x}{1}=\frac{y}{-1}=\frac{z}{1}\) and \(\frac{x-1}{0}=\frac{y+1}{-2}=\frac{z}{1}\) is:

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

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 :

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 :

If the line \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(\quad \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\) intersect, then \(k\) is equal to:

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

Statement 1: The shortest distance between the lines \(\frac{x}{2}=\frac{y}{-1}=\frac{z}{2}\) and \(\frac{x-1}{4}=\frac{y-1}{-2}=\frac{z-1}{4}\) is \(\sqrt{2}\).
Statement 2: The shortest distance between two parallel lines is the perpendicular distance from any point on one of the lines to the other line.

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

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 :

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 \(B(1,6,3)\).

The line \(L\) given by \(\frac{x}{5}+\frac{y}{b}=1\) passes through the point (13,32). The line \(\mathrm{K}\) is parallel to \(L\) and has the equation \(\frac{x}{c}+\frac{y}{3}=1\). Then the distance between \(L\) and \(K\) is

If the straight lines \(\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\frac{x-2}{3}=\frac{y-3}{k} \quad \frac{z-1}{2}\) intersect at a point, then the integer \(\mathrm{k}\) is equal to

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

The angle between the lines \(2 x=3 y=-z\) and \(6 \mathrm{x}=-\mathrm{y}=-4 \mathrm{z}\) is

If the straight lines
\(
x=1+s, y=-3-\lambda s, z=1+\lambda s
\)
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 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

The lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) and \(\frac{x-1}{k}=\frac{y-4}{1}=\frac{z-5}{1}\) are coplanar if

The two lines \(x=a y+\mathrm{b}, z=c y+d\) and \(x=a^{\prime} y+b^{\prime}, z=c^{\prime} y+d^{\prime}\) will be perpendicular, if and only if

If \((2,3,5)\) is one end of diameter of the sphere \(x^2+y^2+z^2\) \(-6 x-12 y-2 z+20=0\), then the coordinates of the other end of the diameter are

The shortest distance between the lines \(\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}\) and \(\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}\) is equal to

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

Consider the lines \(\mathrm{L}_1\) and \(\mathrm{L}_2\) given by
\(L_1: \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-2}{2}\)
\(\mathrm{L}_2: \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-3}{3}\)
A line \(\mathrm{L}_3\) having direction ratios \(1,-1,-2\), intersects \(\mathrm{L}_1\) and \(\mathrm{L}_2\) at the points \(\mathrm{P}\) and \(\mathrm{Q}\) respectively. Then the length of line segment \(P Q\) is

The distance of the point \(\mathrm{P}(4,6,-2)\) from the line passing through the point \((-3,2,3)\) and parallel to a line with direction ratios \(3,3,-1\) is equal to :

The foot of perpendicular of the point \((2,0,5)\) on the line \(\frac{x+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}\) is \((\alpha, \beta, \gamma)\). Then. Which of the following is NOT correct?

The shortest distance between the lines \(x+1=2 y=-\) \(12 \mathrm{z}\) and \(\mathrm{x}=\mathrm{y}+2=6 \mathrm{z}-6\) is