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Exercise 6.5 · Q1

Q.Find the parametric form of vector equation and Cartesian equations of a straight line passing through (5,2,8)(5,2,8) and is perpendicular to the straight lines r⃗=(i^+j^−k^)+s(2i^−2j^+k^)\vec r=(\hat i+\hat j-\hat k)+s(2\hat i-2\hat j+\hat k) and r⃗=(2i^−j^−3k^)+t(i^+2j^+2k^)\vec r=(2\hat i-\hat j-3\hat k)+t(\hat i+2\hat j+2\hat k).

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✓ Free question

A line perpendicular to two given directions must be parallel to their cross product; substitute that direction and the given point into the point-direction formulas.

Step 1. Given directions. b⃗1=(2,−2,1), b⃗2=(1,2,2)\vec b_1=(2,-2,1),\ \vec b_2=(1,2,2).

Step 2. Required direction =b⃗1×b⃗2=\vec b_1\times\vec b_2.

b⃗1×b⃗2=∣i^j^k^2−21122∣=i^(−4−2)−j^(4−1)+k^(4+2)=−6i^−3j^+6k^.\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-2&1\\1&2&2\end{vmatrix}=\hat i(-4-2)-\hat j(4-1)+\hat k(4+2)=-6\hat i-3\hat j+6\hat k.

Simplify by dividing by −3-3: direction (2,1,−2)(2,1,-2).

Step 3. Parametric vector equation through (5,2,8)(5,2,8):

r⃗=(5i^+2j^+8k^)+t(2i^+j^−2k^).\vec r=(5\hat i+2\hat j+8\hat k)+t(2\hat i+\hat j-2\hat k).

Step 4. Cartesian equations.

x−52=y−21=z−8−2.\frac{x-5}{2}=\frac{y-2}{1}=\frac{z-8}{-2}.

✓Final answer

r⃗=(5i^+2j^+8k^)+t(2i^+j^−2k^)\vec r=(5\hat i+2\hat j+8\hat k)+t(2\hat i+\hat j-2\hat k); Cartesian: x−52=y−21=z−8−2\dfrac{x-5}2=\dfrac{y-2}1=\dfrac{z-8}{-2}.

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