RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Maths · 11 questions · 26 marks
This chapter extends coordinate geometry into space, describing lines and planes using direction cosines and direction ratios rather than a single slope. Vector and Cartesian forms of a line's equation turn geometric questions about angles, parallelism and perpendicularity into algebra on direction ratios, while the plane equation and the point-to-plane distance formula let you locate a point relative to any flat surface in three dimensions.
The direction ratios of the line joining the points (1,2,3) and (4,6,15) are proportional to:
Answer
The direction ratios are proportional to 3,4,12. Direction ratios are found by subtracting coordinates: (4-1, 6-2, 15-3) = (3,4,12).
If the direction ratios of a line are 2, -1, 2, then its direction cosines are:
Answer
The direction cosines are (2/3, -1/3, 2/3). The magnitude of (2,-1,2) is √(4+1+4) = √9 = 3. Dividing each ratio by this magnitude gives (2/3, -1/3, 2/3), which does satisfy l²+m²+n² = 4/9+1/9+4/9 = 1.
The distance of the point (1,2,3) from the plane x + 2y + 2z = 13 is:
Answer
The distance is 2/3. Using the formula |ax1+by1+cz1-d|/√(a²+b²+c²) with the plane written as x+2y+2z-13=0: |1+4+6-13|/√(1+4+4) = |-2|/3 = 2/3.
Two lines with direction ratios (1,2,3) and (2,4,6) are:
Answer
The lines are parallel. The second set of direction ratios (2,4,6) is exactly 2 times the first (1,2,3), so the ratios are proportional: 2/1 = 4/2 = 6/3 = 2. This is precisely the condition for two lines to be parallel.
Assertion (A): The lines with direction ratios (1,2,2) and (2,1,-2) are perpendicular. Reason (R): Two lines with direction ratios (a1,b1,c1) and (a2,b2,c2) are perpendicular when a1a2+b1b2+c1c2 = 0.
Answer
Both A and R are true and R is the correct explanation of A. Reason R states the correct general condition for perpendicular lines. For Assertion A: 1(2) + 2(1) + 2(-2) = 2 + 2 - 4 = 0, so applying exactly the condition in R confirms the two lines are perpendicular.
Find the direction cosines of the line joining the points (2,3,4) and (-1,-2,1).
Answer
Direction ratios: (-1-2, -2-3, 1-4) = (-3,-5,-3). Magnitude = √((-3)²+(-5)²+(-3)²) = √(9+25+9) = √43. Direction cosines = (-3/√43, -5/√43, -3/√43).
Find the Cartesian equation of the line passing through the point (1,2,3) with direction ratios 2, 3, -1.
Answer
Using the standard form (x-x1)/a = (y-y1)/b = (z-z1)/c with (x1,y1,z1) = (1,2,3) and (a,b,c) = (2,3,-1): The required equation is (x-1)/2 = (y-2)/3 = (z-3)/(-1).
Find the angle between the lines with direction ratios (2,2,1) and (4,3,12).
Answer
cosθ = (a1a2+b1b2+c1c2) / (√(a1²+b1²+c1²)·√(a2²+b2²+c2²)). Numerator: 2(4) + 2(3) + 1(12) = 8+6+12 = 26. Denominator: √(4+4+1)·√(16+9+144) = √9·√169 = 3×13 = 39. cosθ = 26/39 = 2/3, so θ = cos⁻¹(2/3).
Find the shortest distance between the skew lines r = (î+2ĵ+k̂) + λ(î-ĵ+k̂) and r = (2î-ĵ-k̂) + μ(2î+ĵ+2k̂).
Answer
Here a1=(1,2,1), b1=(1,-1,1), a2=(2,-1,-1), b2=(2,1,2), so a2-a1 = (1,-3,-2). Compute b1×b2, the determinant with rows î ĵ k̂, (1,-1,1), (2,1,2): î: (-1)(2)-(1)(1) = -2-1 = -3. ĵ: -[(1)(2)-(1)(2)] = -(2-2) = 0. k̂: (1)(1)-(-1)(2) = 1+2 = 3. So b1×b2 = (-3,0,3), with |b1×b2| = √(9+0+9) = √18 = 3√2. (a2-a1)·(b1×b2) = (1)(-3) + (-3)(0) + (-2)(3) = -3+0-6 = -9. Shortest distance = |(a2-a1)·(b1×b2)| / |b1×b2| = 9/(3√2) = 3/√2, which can be written as 3√2/2 units.
Find the vector and Cartesian equations of the plane passing through the point (1,1,1) and perpendicular to the line with direction ratios 2,3,-4. Also find the distance of the point (2,3,5) from this plane.
Answer
Since the plane is perpendicular to the given line, the line's direction ratios (2,3,-4) serve as the plane's normal vector n = 2î+3ĵ-4k̂. Vector equation: r·n = a·n, where a = î+ĵ+k̂ is the position vector of the given point. a·n = 2+3-4 = 1, so r·(2î+3ĵ-4k̂) = 1. Cartesian equation: 2(x-1) + 3(y-1) - 4(z-1) = 0, which expands to 2x+3y-4z-1 = 0. Distance of (2,3,5) from this plane: |2(2)+3(3)-4(5)-1|/√(2²+3²+(-4)²) = |4+9-20-1|/√29 = |-8|/√29 = 8/√29.
Read the following and answer the questions that follow: A surveyor recorded two points in hilly terrain: A(1,2,3), the base of a tower, and B(4,6,15), the top of a signal mast. The ground is modelled by the plane x + y + z = 10. (a) Find the direction ratios of the line AB. (b) Find the direction cosines of the line AB. (c) Find the distance of the point A from the plane x + y + z = 10. (d) Does the base point A lie exactly on the plane? Justify using your answer to (c).
Answer
(a) Direction ratios of AB = (4-1, 6-2, 15-3) = (3,4,12). (b) Magnitude = √(9+16+144) = √169 = 13, so direction cosines = (3/13, 4/13, 12/13). (c) Writing the plane as x+y+z-10=0, the distance of A(1,2,3) is |1+2+3-10|/√(1+1+1) = |-4|/√3 = 4/√3 = 4√3/3. (d) No, A does not lie exactly on the plane, since the distance 4√3/3 is nonzero — a point lies on the plane only when its distance to the plane is exactly 0.
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