Class 12 Mathematics Chapter 11 Three-Dimensional Geometry – Extra Questions with Answers

Extra practice questions for Class 12 Maths Chapter 11 (Three-Dimensional Geometry), beyond the textbook. These Class 12 Mathematics Chapter 11 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. What is the relation satisfied by direction cosines l,m,n?
Ans: l²+m²+n²=1.

Q2. Write the vector equation of a line through point a along direction b.
Ans: r=a+λb.

Q3. What is the condition for two lines with direction ratios (a₁,b₁,c₁) and (a₂,b₂,c₂) to be parallel?
Ans: a₁/a₂=b₁/b₂=c₁/c₂.

Q4. Write the Cartesian equation of a plane with normal direction ratios (a,b,c) passing through distance d from origin.
Ans: ax+by+cz=d.

Q5. What is the formula for distance of point (x₁,y₁,z₁) from plane ax+by+cz=d?
Ans: |ax₁+by₁+cz₁−d|/√(a²+b²+c²).

Short Answer Questions (2–3 marks)

Q6. Find the direction cosines of the line joining points (1,2,3) and (4,6,15).
Ans: Direction ratios=(4−1,6−2,15−3)=(3,4,12). Magnitude=√(9+16+144)=√169=13. Direction cosines=(3/13,4/13,12/13).

Q7. Find the angle between lines with direction ratios (1,1,2) and (−1,−1,1).
Ans: cosθ=|(1)(−1)+(1)(−1)+(2)(1)|/[√(1+1+4)√(1+1+1)]=|−1−1+2|/[√6√3]=0/√18=0, so θ=90°.

Q8. Find the distance of the point (2,3,−5) from the plane x+2y−2z−9=0.
Ans: Distance=|1(2)+2(3)−2(−5)−9|/√(1+4+4)=|2+6+10−9|/3=9/3=3 units.

Higher-Order Thinking / Application Questions

Q9. Find the equation of the plane passing through the points (1,1,1), (1,−1,1), and (−7,−3,−5), showing the full determinant method.
Ans: Let the plane pass through A(1,1,1). Find two direction vectors in the plane: AB=B−A=(1−1,−1−1,1−1)=(0,−2,0), and AC=C−A=(−7−1,−3−1,−5−1)=(−8,−4,−6). The normal to the plane is AB×AC, computed via determinant: |i j k; 0 −2 0; −8 −4 −6| = i[(−2)(−6)−(0)(−4)] − j[(0)(−6)−(0)(−8)] + k[(0)(−4)−(−2)(−8)] = i[12−0] − j[0−0] + k[0−16] = 12i+0j−16k = (12,0,−16), simplifiable to (3,0,−4) by dividing by 4. So the plane has normal direction ratios (3,0,−4) and passes through (1,1,1). Its equation is 3(x−1)+0(y−1)−4(z−1)=0, i.e., 3x−4z−3+4=0, i.e., 3x−4z+1=0. Verification: at (1,−1,1): 3(1)−4(1)+1=3−4+1=0 ✓; at (−7,−3,−5): 3(−7)−4(−5)+1=−21+20+1=0 ✓.

Q10. Find the shortest distance between the skew lines r=(i+2j+3k)+λ(i−3j+2k) and r=(4i+5j+6k)+μ(2i+3j+k), explaining why the scalar triple product formula applies.
Ans: Since the two lines have different (non-parallel) direction vectors b₁=(1,−3,2) and b₂=(2,3,1), and do not intersect (they are skew, existing in different planes despite not being parallel), the shortest distance between them is measured along the common perpendicular to both, which requires the scalar triple product formula: d=|(a₂−a₁)·(b₁×b₂)|/|b₁×b₂|. Here a₂−a₁=(4−1,5−2,6−3)=(3,3,3). Compute b₁×b₂: |i j k; 1 −3 2; 2 3 1| = i[(−3)(1)−(2)(3)] − j[(1)(1)−(2)(2)] + k[(1)(3)−(−3)(2)] = i[−3−6] − j[1−4] + k[3+6] = −9i+3j+9k = (−9,3,9). Its magnitude is √(81+9+81)=√171=3√19. The scalar triple product (dot product with (3,3,3)) is (3)(−9)+(3)(3)+(3)(9)=−27+9+27=9. So d=|9|/(3√19)=3/√19 units. This formula works because the numerator projects the connecting vector onto the direction perpendicular to both lines (given by b₁×b₂), isolating exactly the shortest (perpendicular) separation between the two skew lines.

Written by Satish

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