Skip to content
Exercise 6.7 · Q3

Q.Find parametric form of vector equation and Cartesian equations of the plane passing through the points (2,2,1),(1,−2,3)(2,2,1),(1,-2,3) and parallel to the straight line passing through the points (2,1,−3)(2,1,-3) and (−1,5,−8)(-1,5,-8).

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
35% · 57/162 Questions
✓ Free question

The plane through two given points, parallel to a third given direction, has normal equal to the cross product of the in-plane vector AB⃗\vec{AB} and that external direction.

Step 1. In-plane vector. A(2,2,1),B(1,−2,3)A(2,2,1),B(1,-2,3): AB⃗=(1−2,−2−2,3−1)=(−1,−4,2)\vec{AB}=(1-2,-2-2,3-1)=(-1,-4,2).

Step 2. Parallel direction. Through (2,1,−3)(2,1,-3) and (−1,5,−8)(-1,5,-8): (−1−2, 5−1, −8−(−3))=(−3,4,−5)(-1-2,\,5-1,\,-8-(-3))=(-3,4,-5).

Step 3. Normal =AB⃗×=\vec{AB}\times(parallel direction).

∣i^j^k^−1−42−34−5∣=i^(20−8)−j^(5+6)+k^(−4−12)=12i^−11j^−16k^.\begin{vmatrix}\hat i&\hat j&\hat k\\-1&-4&2\\-3&4&-5\end{vmatrix}=\hat i(20-8)-\hat j(5+6)+\hat k(-4-12)=12\hat i-11\hat j-16\hat k.

Step 4. Parametric vector equation through A(2,2,1)A(2,2,1):

r⃗=(2i^+2j^+k^)+s(−i^−4j^+2k^)+t(−3i^+4j^−5k^).\vec r=(2\hat i+2\hat j+\hat k)+s(-\hat i-4\hat j+2\hat k)+t(-3\hat i+4\hat j-5\hat k).

Step 5. Cartesian equation (point-normal, through AA, normal (12,−11,−16)(12,-11,-16)):

12(x−2)−11(y−2)−16(z−1)=0 ⟹ 12x−24−11y+22−16z+16=0 ⟹ 12x−11y−16z+14=0.12(x-2)-11(y-2)-16(z-1)=0\ \Longrightarrow\ 12x-24-11y+22-16z+16=0\ \Longrightarrow\ 12x-11y-16z+14=0.

✓Final answer

Parametric: r⃗=(2i^+2j^+k^)+s(−i^−4j^+2k^)+t(−3i^+4j^−5k^)\vec r=(2\hat i+2\hat j+\hat k)+s(-\hat i-4\hat j+2\hat k)+t(-3\hat i+4\hat j-5\hat k). Cartesian: 12x−11y−16z+14=012x-11y-16z+14=0.

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.