Q.Find the distance of the point (−1,−5,−10) from the point of intersection of the line r=2i^−j^+2k^+λ(3i^+4j^+2k^) and the plane r⋅(i^−j^+k^)=5.
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Vector Equation of a Line
A line is fixed by two pieces of information: one point it passes through and the direction it runs in. The vector equation packages both.
Let a be the position vector of a known point A on the line, and let b be any vector parallel to the line (its direction). For any point P on the line with position vector r, the displacement AP points along the line, so it is a scalar multiple of b: AP=λb. Since r=a+AP,
r=a+λb,λ∈R
How to read it
As the parameter λ runs through all real numbers, r traces every point of the line. At λ=0 you sit at A; positive λ moves one way along b, negative λ the other. Think of a as "where you start" and λb as "how far and which way you walk."
Line through two points
If the line passes through points with position vectors a and b, its direction is b−a, so
r=a+λ(b−a)
Example
The line through A(1,2,−1) parallel to b=2i^−j^+3k^ is
r=(i^+2j^−k^)+λ(2i^−j^+3k^). …
Substitute the line's general point into the plane equation to find λ, get the intersection point, then use the dis …
The required distance is 13 units.
Concept. A point on the line satisfies the plane equation at the intersection; then apply the distance formula.
Why this method. Solving for λ locates the intersection point exactly.
Working. General point on the line: (2+3λ, −1+4λ, 2+2λ). Plane: r⋅(i^−j^+k^)=5: …
Showing the 12 most recent of 19 on this concept.
- KEAM 2026Set eng-2026-04194 marksMCQQ.Consider the straight line r=(5i^+2j^−3k^)+t(4i^+6j^−7k^), t∈R. Which one of the following points, is a point on the straight line? (A) (21,24,−31) (B) (17,20,−22) (C) (1,−4,5) (D) (25,32,−38) (E) (45,66,−36)
›Reveal solutionSolution
A point is on the line iff one parameter t reproduces all three coordinates.
The line is (5+4t,2+6t,−3−7t). Testing option (D) (25,32,−38): 5+4t=25⇒t=5; then 2+6(5)=32 ✓ and −3−7(5)=−38 ✓. …
- KEAM 2026Set eng-2026-04194 marksMCQQ.A straight line passes through the point whose position vector is k^. The straight line also passes through the point of intersection of the lines r=j^+λi^,λ∈R and r=i^+sj^,s∈R. Then the equation of the straight line is (A) r=k^+t(i^+j^−k^),t∈R (B) r=k^+t(i^−j^−k^),t∈R (C) r=k^+t(i^−j^+k^),t∈R (D) r=k^+t(−i^+j^+2k^),t∈R (E) r=k^+t(−i^+2j^−k^),t∈R
›Reveal solutionSolution
Find the intersection point (1,1,0), then take the direction from k^ to it.
Line 1: (λ,1,0); line 2: (1,s,0). They coincide when λ=1, s=1, giving intersection (1,1,0). …
- KEAM 2025Set eng-2025-04254 marksMCQQ.The line 2x+1=4y−4=5z−2 passes through the point (A) (−3,0,−3) (B) (3,0,5) (C) (−3,0,5) (D) (3,4,5) (E) (3,−4,−5)
›Reveal solutionSolution
Write the line in parametric form and test which listed point satisfies it; t=−1 gives (−3,0,−3).
The line 2x+1=4y−4=5z−2=t gives
x=−1+2t,y=4+4t,z=2+5t.
At t=−1:
x=−1−2=−3,y=4−4=0,z=2−5=−3. …
- KEAM 2025Set eng-2025-04254 marksMCQQ.If the line 4x+1=−3y+2=−2z−α passes through the point (−1,−2,−3) then the value of α is (A) 4 (B) −4 (C) 3 (D) −3 (E) −2
›Reveal solutionSolution
Substitute the point into the symmetric equations; the x and y parts force the parameter to 0, giving α=−3.
Line: 4x+1=−3y+2=−2z−α.
At the point (−1,−2,−3):
4x+1=4−1+1=0,−3y+2=−3−2+2=0. …
- KEAM 2025Set eng-2025-04264 marksMCQQ.The Cartesian equation of the line r=(2i^−7j^+11k^)+λ(3i^+7j^−13k^) is (A) 3x−2=7y+7=−13z−11 (B) 3x−2=7y−7=13z−11 (C) 3x+2=7y−7=−13z+11 (D) 3x+2=7y+7=−13z−11 (E) 3x+2=13y=−7z−11
›Reveal solutionSolution
The vector line r=a+λd becomes d1x−a1=d2y−a2=d3z−a3; here a=(2,−7,11), d=(3,7,−13).
Writing components: x=2+3λ, y=−7+7λ, z=11−13λ. Eliminating λ, …
- KEAM 2025Set eng-2025-04264 marksMCQQ.Which one of the following is a point on the straight line r=(13i^−14j^+23k^)+λ(5i^−7j^−9k^), λ∈R (A) (13,−14,−23) (B) (5,−7,−9) (C) (23,−28,7) (D) (23,−28,5) (E) (13,14,23)
›Reveal solutionSolution
Points on the line are (13+5λ,−14−7λ,23−9λ); λ=2 gives (23,−28,5).
Parametrize: x=13+5λ,y=−14−7λ,z=23−9λ. Try λ=2: x=23,y=−28,z=23−18=5, giving (23,−28,5), which …
- KEAM 2025Set eng-2025-04274 marksMCQQ.The equation of the line passing through (0,0,1) and (1,1,0) is (A) r=k^+λ(i^+j^−k^),λ∈R (B) r=j^+λ(i^−j^+k^),λ∈R (C) r=i^+λ(i^+j^+k^),λ∈R (D) r=i^+j^+λ(i^−j^−k^),λ∈R (E) r=i^+j^+k^+λ(i^+j^−k^),λ∈R
›Reveal solutionSolution
The direction vector is (1,1,−1) and the point (0,0,1)=k^, so r=k^+λ(i^+j^−k^).
The line passes through (0,0,1) and (1,1,0). Its direction is
(1,1,0)−(0,0,1)=(1,1,−1).
Using the point (0,0,1)=k^: …
- KEAM 2025Set eng-2025-04294 marksMCQQ.Which one of the following points lies on the line r=(i^+2j^−3k^)+t(4i^+5j^−7k^), t∈R? (A) (9,12,−15) (B) (9,15,12) (C) (12,9,−17) (D) (9,12,−17) (E) (−9,−12,17)
›Reveal solutionSolution
Points on the line are (1+4t,2+5t,−3−7t); t=2 gives (9,12,−17), matching option (D). …
- KEAM 2024Set eng-2024-06054 marksMCQQ.The common point of the two straight lines r=(i^−2j^+3k^)+s(2i^+j^+k^) and r=(−i^+2j^+7k^)+t(i^+j^+k^), t,s∈R is (A) (11,8,−3) (B) (−11,−8,−3) (C) (11,−8,3) (D) (11,−8,−3) (E) (9,8,−3)
›Reveal solutionSolution
Set the parametric points equal and solve for s,t, then substitute.
Line 1 point: (1+2s,−2+s,3+s); Line 2 point: (−1+t,2+t,7+t).
From the y- and z-equations: −2+s=2+t and 3+s=7+t both give s−t=4.
From the x-equation: 1+2s=−1+t⇒2s−t=−2. …
- KEAM 2024Set eng-2024-06064 marksMCQQ.The point of intersection of the lines 2x−3=2y−2=1z−6 and 3x−2=2y−4=3z−1 is (A) (3,4,3) (B) (7,6,6) (C) (4,3,3) (D) (10,11,10) (E) (11,10,10)
›Reveal solutionSolution
Parametrize both lines and solve for the shared point.
Line 1: x=3+2t, y=2+2t, z=6+t. Line 2: x=2+3s, y=4+2s, z=1+3s.
From y: 2+2t=4+2s⇒t−s=1. From x: 3+2t=2+3s⇒2t−3s=−1. Solving: t=4, s=3. …
- KEAM 2024Set eng-2024-06074 marksMCQQ.The line joining the points (2,2,2) and (6,6,6) meets the line 3x−1=2y−2=−1z−5 at the point (A) (1,1,1) (B) (2,2,2) (C) (3,3,3) (D) (4,4,4) (E) (6,6,6)
›Reveal solutionSolution
The line through (2,2,2) and (6,6,6) is x=y=z=t. Substituting into 3x−1=2y−2=−1z−5: 3t−1=2t−2⇒t=4, and −14−5=1=34−1 checks. Point (4,4,4).
The two given points (2,2,2) and (6,6,6) both satisfy x=y=z, so points on this line are (t,t,t).
Substitute into the second line: …
- KEAM 2024Set eng-2024-06084 marksMCQQ.Equation of the line parallel to the line 2x−2=3y−2=−2z−1 and passing through the point (3,2,−1) is (A) 2x−3=3y−2=2z+1 (B) 2x+3=3y+2=−2z−1 (C) 2x−3=3y−2=−2z−1 (D) 2x−3=3y−2=−2z+1 (E) 2x+3=3y+2=−2z+1
›Reveal solutionSolution
A parallel line has the same direction ratios (2,3,−2); passing through (3,2,−1) gives 2x−3=3y−2=−2z+1. …
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