Q.Find the vector equation of the line which passes through the point (3,2,1) and is parallel to the vector 2i^+2j^−3k^.
Concept understanding — Vector Equation of a Line through a Point and Parallel to a Vector
A line in space is fully determined once we know one point on it and the direction it runs in. If A is a known point with position vector aˉ, and bˉ is a vector parallel to the line, then any other point P(rˉ) on the line satisfies AP=λbˉ for some scalar λ, because AP must itself run in the direction of bˉ. Since AP=rˉ−aˉ, this gives the vector equation rˉ=aˉ+λbˉ. Here λ is a free real parameter — every real value of λ produces one point on the line, and every point on the line corresponds to exactly one value of λ, so this is called the parametric form of the line's vector equation. The same relationship can be written without a parameter as (rˉ−aˉ)×bˉ=0ˉ, called the non-parametric form, since two parallel vectors always have a zero cross product. This idea is the workhorse behind many line problems: even when the direction vector bˉ is not handed to us directly, it can often be built — from a set of direction ratios, from the vector joining two other points, or as a cross product of two vectors the line must be perpendicular to (since bˉ×cˉ is always perpendicular to both bˉ and cˉ) — and then substituted into this same formula.
[!TLDR] Convert the point (3,2,1) to a position vector and pair it with the given direction vector.
[!ANSWER] r=(3i^+2j^+k^)+λ(2i^+2j^−3k^).
The point (3,2,1) has position vector a=3i^+2j^+k^, and the line is parallel to b=2i^+2j^−3k^.
Using the standard form r=a+λb:
[!ANSWER] r=(3i^+2j^+k^)+λ(2i^+2j^−3k^).
Convert the given point to its position vector and substitute into r=a+λb together with the given parallel vector.
Mis-ordering the coordinates (3,2,1) while writing the position vector; sign error copying −3k^ from the given direction vector.
- CBSE 2025Set ANNUAL2 marksMCQQ.The vector equation of the line passing through the point having position vector 4i^−j^+2k^ and parallel to vector −2i^−j^+k^ is given by ____.(a) (4i^−j^−2k^)+λ(−2i^−j^+k^)(b) (4i^−j^+2k^)+λ(2i^−j^+k^)(c) (4i^−j^+2k^)+λ(−2i^−j^−k^)(d) (4i^−j^+2k^)+λ(−2i^−j^+k^)
›Reveal solutionSolution
Vector equation of a line through a point A(aˉ) parallel to bˉ is rˉ=aˉ+λbˉ.
The line passes through the point with position vector aˉ=4i^−j^+2k^ and is parallel to bˉ=−2i^−j^+k^.
The vector equation of a line through a fixed point aˉ and parallel to bˉ is
rˉ=aˉ+λbˉ
Substituting,
rˉ=(4i^−j^+2k^)+λ(−2i^−j^+k^)
This matches option (d) exactly (options (a), (b), (c) alter a sign in either the point or the direction vector).
✓Final answerrˉ=(4i^−j^+2k^)+λ(−2i^−j^+k^), option (d).
- CBSE 2024Set ANNUAL2 marksQ.Find the vector equation of the line passing through the point having position vector 4i^−j^+2k^ and parallel to the vector −2i^−j^+k^.
›Reveal solutionSolution
Vector equation: (point) + λ(direction).
Line through position vector aˉ=4i^−j^+2k^, parallel to bˉ=−2i^−j^+k^.
Vector equation: rˉ=aˉ+λbˉ=(4i^−j^+2k^)+λ(−2i^−j^+k^), λ∈R
✓Final answerrˉ=(4i^−j^+2k^)+λ(−2i^−j^+k^)
- CBSE 2018Set ANNUAL2 marksQ.Find the vector equation of the line which passes through the point with position vector 4i^−j^+2k^ and is in the direction of −2i^+j^+k^.
›Reveal solutionSolution
Vector equation of a line through a point aˉ in the direction bˉ is rˉ=aˉ+λbˉ.
The line passes through the point with position vector aˉ=4i^−j^+2k^ and has direction vector bˉ=−2i^+j^+k^.
The vector equation of a line through a given point with position vector aˉ along direction bˉ is:
rˉ=aˉ+λbˉ,λ∈R
Substituting:
rˉ=(4i^−j^+2k^)+λ(−2i^+j^+k^)
✓Final answerrˉ=(4i^−j^+2k^)+λ(−2i^+j^+k^)
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