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Three Dimensional Geometry Set-1 MCQ Online Practice Test | Class 12 | Quiz Tweek

Three Dimensional Geometry Set-1

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

Three Dimensional Geometry is crucial in both board and competitive exams because it builds the core skill of working with planes, lines, spheres, and their relative positions in 3D. The chapter directly trains you for computing distances, angles, intersections, and equations using vector and analytic geometry—exactly the types of problems that appear frequently in CBSE and JEE/NEET.

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Q1. The planes $2x-3y+6z-5=0$ and $4x-6y+12z-11=0$ are given. The distance between them is

(A)

$\dfrac{1}{7}$

(B)

$\dfrac{1}{28}$

(C)

$\dfrac{1}{14}$

(D)

$\dfrac{6}{7}$

Q2. Let the line $L$ be given by $\dfrac{x-1}{1}=\dfrac{y-2}{-2}=\dfrac{z-3}{2}$ . The point on $L$ closest to the origin and the distance from the origin to $L$ are

(A)

$\left(1,2,3\right),\ \sqrt{14}$

(B)

$\left(\dfrac{2}{3},\dfrac{8}{3},\dfrac{7}{3}\right),\ \sqrt{13}$

(C)

$\left(\dfrac{4}{3},\dfrac{4}{3},\dfrac{11}{3}\right),\ \sqrt{17}$

(D)

$\left(2,0,5\right),\ \sqrt{29}$

Q3. Find the shortest distance between the skew lines $L_1:\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{-1}$ and $L_2:\dfrac{x+2}{1}=\dfrac{y-1}{-2}=\dfrac{z+4}{4}$ .

(A)

$\dfrac{14}{\sqrt{230}}$

(B)

$\dfrac{28}{\sqrt{460}}$

(C)

$\dfrac{7}{\sqrt{230}}$

(D)

$\dfrac{28}{\sqrt{230}}$

Q4. Assertion (A): For two skew lines $L_1,L_2$ with direction vectors $\mathbf{d}_1,\mathbf{d}_2$ , the shortest distance between them equals the absolute value of the projection of any vector joining a point on $L_1$ to a point on $L_2$ onto the direction of $\mathbf{n}=\mathbf{d}_1\times\mathbf{d}_2$ .
Reason (R): The shortest segment joining two skew lines is perpendicular to both lines and hence is parallel to $\mathbf{d}_1\times\mathbf{d}_2$ .

(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.

Q5. Let $P_1:x+y+z=3$ and $P_2:2x-y+3z=4$ . Let $L$ be the line of intersection of $P_1$ and $P_2$ . The point on $L$ nearest to the origin and the distance from the origin to $L$ are

(A)

$\left(\dfrac{7}{3},\dfrac{2}{3},0\right),\ \dfrac{\sqrt{53}}{3}$

(B)

$\left(1,1,1\right),\ \sqrt{3}$

(C)

$\left(-\dfrac{1}{3},\dfrac{4}{3},2\right),\ \dfrac{\sqrt{53}}{3}$

(D)

$\left(\dfrac{11}{3},\dfrac{1}{3},-1\right),\ \dfrac{\sqrt{131}}{3}$

Q6. Let the line pass through $A(1,2,3)$ and $B(4,0,-1)$ . Let the plane pass through $P(2,1,3)$ , $Q(0,1,1)$ and $R(1,3,0)$ . The acute angle between the line $AB$ and the plane $PQR$ is

(A)

$\arccos\!\left(\dfrac{9}{\sqrt{87}}\right)\text{ (approximately }15.1^\circ\text{)}$

(B)

$\displaystyle \arcsin\!\left(\dfrac{9}{\sqrt{87}}\right)\text{ (approximately }74.9^\circ\text{)}$

(C)

$\arccos\!\left(\dfrac{3}{\sqrt{29}}\right)\text{ (approximately }56.1^\circ\text{)}$

(D)

$\displaystyle \arcsin\!\left(\dfrac{3}{\sqrt{29}}\right)\text{ (approximately }33.8^\circ\text{)}$

Q7. Lines $L_1$ and $L_2$ are given by $L_1:\ (x,y,z)=(1,0,1)+t(2,-1,1)$ and $L_2:\ (x,y,z)=(0,1,2)+s(1,1,-2)$ . The coordinates of the point on $L_1$ closest to $L_2$ are

(A)

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

(B)

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

(C)

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

(D)

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

Q8. Find the equation of the plane which passes through the line of intersection of the planes $x-y+2z-3=0$ and $2x+y-z+1=0$ and is perpendicular to the plane $x+3y-2z+4=0$ .

(A)

$19x-y+8z-15=0$

(B)

$19x+y+8z-15=0$

(C)

$7x-y+3z-5=0$

(D)

$19x-y+8z+15=0$

Q9. The sphere $(x-4)^2+(y+1)^2+(z-3)^2=25$ intersects the plane $2x-3y+6z-1=0$ in a circle. The center and radius of this circle are

(A)

center $\,(4,-1,3)$ , radius $3$

(B)

center $\!\left(\dfrac{20}{7},-\dfrac{5}{7},\dfrac{3}{7}\right)$ , radius $3$

(C)

center $\!\left(\dfrac{20}{7},\dfrac{5}{7},-\dfrac{3}{7}\right)$ , radius $3$

(D)

center $\!\left(\dfrac{20}{7},\dfrac{5}{7},-\dfrac{3}{7}\right)$ , radius $\sqrt{15}$

Q10. The line $l$ passes through $A(1,2,-1)$ with direction vector $(2,-1,2)$ . It is reflected in the plane $\pi:\ x-y+2z+3=0$ to give the line $l'$ . The equation of the reflected line $l'$ is

(A)

$\displaystyle \frac{x-1}{1}=\frac{y-2}{-4}=\frac{z+1}{8}$

(B)

$\displaystyle (x,y,z)=(1,2,-1)+t(-1,4,-8)$

(C)

$\displaystyle (x,y,z)=(1,2,-1)+t(1,-4,8)$

(D)

$\displaystyle (x,y,z)=(1,2,-1)+t(-1,4,8)$

Q11. A point $P(1,2,3)$ is reflected across the plane $2x-y+2z-3=0$ to a point $P'$ . The coordinates of $P'$ are:

(A)

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

(B)

$\left(-\dfrac{1}{3},\dfrac{8}{3},\dfrac{5}{3}\right)$

(C)

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

(D)

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

Q12. Find the shortest distance between the skew lines $L_1:\;\mathbf r=(1,0,2)+t(1,1,-1)$ and $L_2:\;\mathbf r=(2,1,0)+s(2,-1,1)$ .

(A)

$\dfrac{1}{2}$

(B)

$1$

(C)

$\dfrac{3}{\sqrt{10}}$

(D)

$\dfrac{1}{\sqrt{2}}$

Q13. A sphere passes through $P(1,2,3)$ and $Q(3,0,1)$ and its centre lies on the line given by $\dfrac{x-1}{1}=\dfrac{y-1}{1}=\dfrac{z}{1}$ . The equation of the sphere is:

(A)

$(x-2)^2+(y-2)^2+(z-1)^2=5$

(B)

$(x-2)^2+(y-2)^2+(z-1)^2=6$

(C)

$(x-1)^2+(y-1)^2+(z-1)^2=5$

(D)

$(x-3)^2+(y-1)^2+(z-2)^2=5$

Q14. Let $L$ be the line through $A(1,2,3)$ perpendicular to the plane $\pi:\;x+2y+2z-7=0$ . Let $M$ be the line of intersection of $\pi$ with the plane $\sigma:\;2x-y+z-4=0$ . The acute angle between $L$ and $M$ is:

(A)

$30^\circ$

(B)

$60^\circ$

(C)

$90^\circ$

(D)

$45^\circ$

Q15. The planes $\pi_1:\;x+y+z-3=0$ and $\pi_2:\;2x-y+z-1=0$ intersect in the line $L$ . Let $P(3,0,0)$ . The point on $L$ nearest to $P$ is:

(A)

$\left(\dfrac{11}{7},\dfrac{25}{14},-\dfrac{5}{14}\right)$

(B)

$\left(\dfrac{4}{3},\dfrac{5}{3},0\right)$

(C)

$\left(\dfrac{13}{7},\dfrac{23}{14},\dfrac{5}{14}\right)$

(D)

$\left(\dfrac{11}{7},\dfrac{25}{14},\dfrac{5}{14}\right)$

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