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Three Dimensional Geometry Set-6 MCQ Online Practice Test | Class 12 | Quiz Tweek

Three Dimensional Geometry Set-6

Practice Important MCQs for Three Dimensional Geometry — Mathematics, Class 12.

Three Dimensional Geometry is crucial in CBSE boards and competitive exams because it builds the foundation for understanding planes, lines, distances, and angles in 3D space. Most questions are based on standard vector/coordinate geometry results (direction vectors, cross products, dot products, and distance formulas), so mastering this chapter directly improves accuracy and speed in higher-level problems across Class 12, JEE, and NEET.

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Q1. Find the shortest distance from the point $P(1,2,-1)$ to the plane $2x-3y+6z-7=0$ .

(A)

$\displaystyle \frac{17}{3}$

(B)

$\displaystyle \frac{17}{7}$

(C)

$\displaystyle \frac{7}{17}$

(D)

$\displaystyle \frac{5}{7}$

Q2. Find the equation of the plane which passes through the line of intersection of the planes $x+2y+3z-4=0$ and $2x-y+z+1=0$ and is perpendicular to the plane $x+y+z-3=0$ .

(A)

$5x-5y+7=0$

(B)

$5x-5y-7=0$

(C)

$x-y+7=0$

(D)

$5x+5y+7=0$

Q3. The lines $L_1:\;\dfrac{x-1}{1}=\dfrac{y-0}{2}=\dfrac{z-1}{-1}$ and $L_2:\;\dfrac{x-2}{2}=\dfrac{y-1}{-1}=\dfrac{z-0}{1}$ are skew. The shortest distance between $L_1$ and $L_2$ equals

(A)

$\displaystyle \frac{1}{\sqrt7}$

(B)

$\displaystyle \frac{3}{\sqrt{14}}$

(C)

$0$

(D)

$\displaystyle \frac{3}{\sqrt{35}}$

Q4. Let $L_1: \mathbf r=(1,0,0)+t(1,-1,2)$ and $L_2: \mathbf r=(0,2,1)+s(2,1,-1)$ . Points $P$ on $L_1$ and $Q$ on $L_2$ are chosen so that $PQ$ is the shortest segment joining the two lines. Then

(A)

$P\!\left(\frac{4}{5},-\frac{1}{5},-\frac{2}{5}\right),\;Q\!\left(\frac{2}{5},\frac{11}{5},\frac{4}{5}\right)$

(B)

$P\!\left(\frac{4}{5},\frac{1}{5},\frac{2}{5}\right),\;Q\!\left(\frac{2}{5},\frac{11}{5},-\frac{4}{5}\right)$

(C)

$P\!\left(\frac{4}{5},\frac{1}{5},-\frac{2}{5}\right),\;Q\!\left(\frac{2}{5},\frac{11}{5},\frac{4}{5}\right)$

(D)

$P\!\left(\frac{4}{5},\frac{1}{5},-\frac{2}{5}\right),\;Q\!\left(\frac{2}{5},\frac{11}{5},-\frac{4}{5}\right)$

Q5. Assertion (A): If two non-parallel planes are each perpendicular to a third plane, then their line of intersection is perpendicular to that third plane.
Reason (R): For a plane $ax+by+cz+d=0$ the vector $(a,b,c)$ is a normal to the plane and gives its direction ratios.

(A)

Both A and R are true and R is the correct explanation of A.

(B)

Both A and R are true but R is NOT the correct explanation of A.

(C)

A is true but R is false.

(D)

A is false but R is true.

Q6. Find the perpendicular distance from the point $P(1,2,3)$ to the plane $2x-3y+6z-12=0$ .

(A)

$\dfrac{1}{7}$

(B)

$\dfrac{6}{7}$

(C)

$\dfrac{2}{7}$

(D)

$\dfrac{2}{49}$

Q7. The foot of the perpendicular from the point $P(4,-1,3)$ to the plane $x-2y+2z+3=0$ is

(A)

$\left(\dfrac{7}{3},\dfrac{7}{3},-\dfrac{1}{3}\right)$

(B)

$\left(\dfrac{5}{3},\dfrac{1}{3},\dfrac{2}{3}\right)$

(C)

$\left(\dfrac{4}{3},-\dfrac{2}{3},\dfrac{5}{3}\right)$

(D)

$\left(\dfrac{8}{3},\dfrac{2}{3},-\dfrac{2}{3}\right)$

Q8. Let the line $L$ be given by $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-2}{1}$ . Find the equation of the plane which contains $L$ and is perpendicular to the plane $3x-y+z-4=0$ .

(A)

$y+z+1=0$

(B)

$y+z-1=0$

(C)

$x+y+z-2=0$

(D)

$y-z-1=0$

Q9. The shortest distance between the two skew lines
$L_1:\dfrac{x-1}{2}=\dfrac{y+2}{-1}=\dfrac{z-3}{1}$ and $L_2:\dfrac{x+2}{1}=\dfrac{y-1}{2}=\dfrac{z+1}{-2}$ is

(A)

$\dfrac{1}{2}$

(B)

$\dfrac{3}{2}$

(C)

$\dfrac{\sqrt{3}}{2}$

(D)

$\dfrac{\sqrt{2}}{2}$

Q10. Let $L_1$ be given by $(x,y,z)=(t,t,0)$ and $L_2$ by $(x,y,z)=(s,-s,1)$ for real parameters $t,s$ . If $M$ is the midpoint of the segment joining a point on $L_1$ to a point on $L_2$ , the locus of $M$ and its distance from the origin are

(A)

Plane $z=\dfrac{1}{2}$ , distance $\dfrac{1}{2}$

(B)

Plane $z=1$ , distance $1$

(C)

Plane $y=0$ , distance $0$

(D)

Plane $x=\dfrac{1}{2}$ , distance $\dfrac{1}{2}$

Q11. The line $L$ is given by

\mathbf r=\langle1,2,3\rangle+\lambda\langle2,-1,2\rangle.

The point $P$ is $(3,1,0)$ . The perpendicular distance from $P$ to $L$ is

(A)

$\displaystyle \frac{3\sqrt5}{2}$

(B)

$4$

(C)

$\displaystyle \frac{5\sqrt5}{3}$

(D)

$\displaystyle \frac{\sqrt{35}}{3}$

Q12. Let

\mathbf r=\langle1,0,2\rangle+\lambda\langle1,2,-1\rangle,\qquad L_2:\ \mathbf r=\langle2,1,0\rangle+\mu\langle2,-1,1\rangle.

.
The shortest distance between $L_1$ and $L_2$ equals

(A)

$\displaystyle \frac{8}{\sqrt{35}}$

(B)

$\displaystyle \frac{4}{\sqrt{14}}$

(C)

$\displaystyle \frac{\sqrt{35}}{8}$

(D)

$\displaystyle \frac{2\sqrt{7}}{7}$

Q13. The sphere $S$ has equation

x^2+y^2+z^2-4x+2y-6z+1=0

and it meets the plane

2x-y+2z=3

in a circle. The radius of that circle is

(A)

$\displaystyle \frac{10\sqrt2}{3}$

(B)

$\displaystyle \frac{\sqrt{53}}{3}$

(C)

$\displaystyle \frac{\sqrt{13}}{3}$

(D)

$\displaystyle \frac{8}{3}$

Q14. For the skew lines

L_1:\ \mathbf r=\langle1,1,0\rangle+\lambda\langle1,2,3\rangle,\qquad L_2:\ \mathbf r=\langle2,0,1\rangle+\mu\langle2,-1,1\rangle,

find the equation of the common perpendicular and its length.

(A)

\mathbf r=\langle1,1,0\rangle+t\langle1,0,1\rangle,\qquad \text{length } \frac{\sqrt3}{3}

(B)

\mathbf r=\langle1,0,1\rangle+t\langle-1,-1,1\rangle,\qquad \text{length } \frac{\sqrt3}{3}

(C)

\mathbf r=\langle1,1,0\rangle+t\langle-1,-1,1\rangle,\qquad \text{length } \frac{\sqrt6}{6}

(D)

\mathbf r=\langle1,1,0\rangle+t\langle-1,-1,1\rangle,\qquad \text{length } \frac{\sqrt3}{3}

Q15. Let

L:\ \mathbf r=\langle1,2,0\rangle+t\langle1,-1,2\rangle

and points $A(3,0,1),\ B(1,4,-1)$ . Find all plane(s) that contain $L$ and are equidistant from $A$ and $B$ .

(A)

Both planes

3x+y-z-5=0\quad\text{and}\quad 2y+z-4=0

(B)

Only

3x+y-z-5=0

(C)

Only

2y+z-4=0

(D)

No such plane exists

...and 5 more challenging questions available in the interactive simulator.

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