Q.Find the Cartesian equation of a line passing through the Points (3,-2,-5) and (3,-2,6).
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Symmetric Form To Vector
Symmetric Form to Vector Form: The Intuition
There are two ways to describe a straight line in space. One says: "Start at this point, and move in this direction" — the vector form, like a starting location plus a compass bearing. The other says: "For every step east, I take two north and three up" — the symmetric form, giving the ratio of movement between coordinates.
Symmetric form:
ax−x0=by−y0=cz−z0
Vector form:
r=⟨x0,y0,z0⟩+t⟨a,b,c⟩
They say the same thing — the symmetric form is just the vector form written as equal ratios.
How to Convert Symmetric Form to Vector Form
Step 1: Read off the fixed point. The numerators x0,y0,z0 give a point on the line. In
3x−2=−4y+1=2z−5
the point is (2,−1,5). Watch the signs: y+1 means y−(−1), so y0=−1.
Step 2: Read off the direction vector. The denominators give the direction: ⟨3,−4,2⟩.
Step 3: Write the vector form.
r=⟨2,−1,5⟩+t⟨3,−4,2⟩
If a denominator is zero, that coordinate is constant. For 2x−1=0y+3=5z−4, y=−3 always — the line is parallel to the xz-plane. In vector form: r=⟨1,−3,4⟩+t⟨2,0,5⟩.
Why It Works
From the vector form,
x=x0+at,y=y0+bt,z=z0+ct
Solve each for t:
t=ax−x0,t=by−y0,t=cz−z0
Since all three equal the same t, they equal each other — that's the symmetric form. The conversion just reads this chain backwards.
The symmetric form needs a,b,c all non-zero. If a component is zero you can't divide by it — but the conversion still works: that coordinate is constant, and you omit it from the chain.
--- …
The direction ratios come from the coordinate differences; here two are zero. The OR uses cosθ=∣n1∣∣n2∣∣n1⋅n2∣ for the angle between planes. …
Main: the line through (3,−2,−5),(3,−2,6) is x=3, y=−2 (direction along the z-axis). OR: angle between planes =cos−1735.
Main part. Direction ratios =(3−3, −2−(−2), 6−(−5))=(0,0,11). The symmetric (Cartesian) form is
0x−3=0y+2=11z+5,
which means x=3, y=−2 with z free (a line parallel to the z-axis).
OR part. Planes 3x−6y+2z=7 and 2x+2y−2z=5 have normals n1=(3,−6,2), n2=(2,2,−2). …
- CBSE 2026Set ANNUAL1 markQ.The vector equation of the line 7x−5=7y+4=26−z is ..........
›Reveal solutionSolution
Rewrite the Cartesian equation so all terms have the form direction ratiocoordinate−point, then read off the point and direction vector.
Given: 7x−5=7y+4=26−z
Rewrite the third term: 26−z=2−(z−6)=−2z−6
So: 7x−5=7y−(−4)=−2z−6
…
- CBSE 2026Set ANNUAL1 markMCQQ.Vector equation of the line (x-5)/-4 = (y-3)/5 = (z+3)/-8 is:(a) vector r = 4i - 5j - 8k + μ(5i + 3j - 3k)(b) vector r = -4i + 5j + 8k + μ(5i + 3j - 3k)(c) vector r = 5i + 3j - 3k + μ(4i - 5j - 8k)(d) vector r = 5i + 3j - 3k + μ(-4i + 5j - 8k)
›Reveal solutionSolution
The Cartesian symmetric form ax−x1=by−y1=cz−z1 directly gives a point (x1,y1,z1) on the line and direction ratios ⟨a,b,c⟩, from which the vector equation follows.
The given line is:
−4x−5=5y−3=−8z+3
Comparing with ax−x1=by−y1=cz−z1, the point on the line is (5,3,−3) and direction ratios are ⟨−4,5,−8⟩.
…
- CBSE 2026Set ANNUAL1 markQ.Write the answer in one word/sentence: Write the equation of a line in Cartesian form which passes through a point (x1,y1,z1) and whose direction cosines are l,m,n.
›Reveal solutionSolution
Cartesian equation of a line through a point with given direction cosines.
A line through (x1,y1,z1) with direction cosines l,m,n has Cartesian equation
…
- CBSE 2025Set E1 markMCQQ.Through which of the following points does the line 12x−11=13y−12=14z+13 pass?(a) 11,12,13(b) 11,12,−13(c) 12,13,14(d) −11,−12,13
›Reveal solutionSolution
The line passes through (x1,y1,z1) where each numerator vanishes: (11,12,−13).
Write 12x−11=13y−12=14z+13. Setting each fraction to 0 gives x=11, y=12, …
- CBSE 2025Set ANNUAL1 markMCQQ.The Cartesian equation of a line is 2x−1=3y+2=−1z−5. Its vector equation is(a) r=(−i^+2j^−5k^)+λ(2i^+3j^−k^)(b) r=(2i^+3j^−k^)+λ(i^−2j^+5k^)(c) r=(i^−2j^+5k^)+λ(2i^+3j^−k^)(d) none of these
›Reveal solutionSolution
Read the fixed point and direction ratios straight off the Cartesian form and assemble the vector equation r = a + λb.
The Cartesian equation 2x−1=3y+2=−1z−5 is of the standard form ax−x1=by−y1=cz−z1, which passes through (x1,y1,z1)=(1,−2,5) with direction ratios (a,b,c)=(2,3,−1).
The corresponding vector equation is:
…
- CBSE 2025Set ANNUAL1 markMCQQ.Vector equation of the line (x−5)/−4 = (y−3)/5 = (z+3)/−8 is:(a) r⃗ = 4î − 5ĵ − 8k̂ + μ(5î + 3ĵ − 3k̂)(b) r⃗ = −4î + 5ĵ + 8k̂ + μ(5î + 3ĵ − 3k̂)(c) r⃗ = 5î + 3ĵ − 3k̂ + μ(4î − 5ĵ − 8k̂)(d) r⃗ = 5î + 3ĵ − 3k̂ + μ(−4î + 5ĵ − 8k̂)
›Reveal solutionSolution
Read the point and direction ratios straight off the Cartesian symmetric form, then write the vector equation r=a+μb.
The Cartesian equation is −4x−5=5y−3=−8z−(−3).
Comparing with ax−x1=by−y1=cz−z1: the line passes through the point (5,3,−3) and has direction ratios (−4,5,−8).
…
- CBSE 2023Set E1 markMCQQ.Through which of the following points does the straight line 101x−100=102y−99=103z−98 pass?(a) (101,102,103)(b) (98,99,100)(c) (100,99,98)(d) (99,100,101)
›Reveal solutionSolution
The point (x0,y0,z0) in ax−x0=⋯ lies on the line, here (100,99,98).
The symmetric form 101x−100=102y−99=103z−98 passes through the point (100,99,98), obtained by setting each fraction equal to 0.
…
- CBSE 2023Set ANNUAL1 markQ.The cartesian equation of a line is 3x−5=7y+4=2z−6. Write its vector form.
›Reveal solutionSolution
A cartesian line ax−x0=by−y0=cz−z0 passes through (x0,y0,z0) with direction (a,b,c); write this as a vector equation.
Given 3x−5=7y+4=2z−6, the line passes through the point (5,−4,6) with direction ratios (3,7,2).
…
- CBSE 2020Set ANNUAL1 markMCQQ.The point through which the straight line 22x−3=−52y+3=2z passes is(a) (23,2−3,0)(b) (2−3,23,0)(c) (23,23,0)(d) none of these
›Reveal solutionSolution
Rewrite each fraction in the form (variable − constant)/(direction ratio) to read off the point the line passes through.
22x−3=−52y+3=2z
Rewrite each term: 22x−3=22(x−23)=x−23=1x−23
−52y+3=−52(y+23)=−5/2y−(−23)
2z=2z−0 …
- CBSE 2019Set ANNUAL1 markQ.The certesian equation of a line is 2x−1=3y−2=4z−3. Write its vector form.
›Reveal solutionSolution
read off the point and direction ratios from the cartesian form
2x−1=3y−2=4z−3 passes through (1,2,3) with direction ratios (2,3,4).
…
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.