Question 9 of 25intermediate🔧 ApplyShort Answer4 marks

Find the equation of the line which passes through the point (1,2,3)(1, 2, 3) and is parallel to the vector 3i^+2j^−2k^3\hat{i} + 2\hat{j} - 2\hat{k}.

Correct Answer

Vector Form: Let the position vector of the given point (1, 2, 3) be a⃗ = î + 2ĵ + 3k̂. The line is parallel to the vector b⃗ = 3î + 2ĵ - 2k̂. Therefore, the vector equation of the line is: r⃗ = (î + 2ĵ + 3k̂) + λ(3î + 2ĵ - 2k̂)

Cartesian Form: Let r⃗ be the position vector of any point P(x, y, z) on the line. Then, xî + yĵ + zk̂ = (î + 2ĵ + 3k̂) + λ(3î + 2ĵ - 2k̂)

Equating the coefficients of î, ĵ, k̂: x = 1 + 3λ, y = 2 + 2λ, z = 3 - 2λ

Eliminating λ, we get: (x-1)/3 = (y-2)/2 = (z-3)/(-2)

This is the required Cartesian equation of the line.

Exercise: EXERCISE 11.2 | Q: 4 | (Page Page 13)
For More Understanding

Explanation

This question follows the same pattern as Example 6 in the textbook. The student needs to identify that the position vector of point (1,2,3) gives the constant term, and the parallel vector gives the direction ratios. Both vector and Cartesian forms should be provided for a complete answer worth 4 marks.

Solution Steps

  1. Step 1: Identify the position vector of the given point: a⃗ = î + 2ĵ + 3k̂ for point (1, 2, 3)

  2. Step 2: Identify the direction vector: b⃗ = 3î + 2ĵ - 2k̂

  3. Step 3: Write the vector equation: r⃗ = a⃗ + λb⃗ = (î + 2ĵ + 3k̂) + λ(3î + 2ĵ - 2k̂)

  4. Step 4: For Cartesian form, equate coefficients: x = 1 + 3λ, y = 2 + 2λ, z = 3 - 2λ

  5. Step 5: Eliminate λ to get: (x-1)/3 = (y-2)/2 = (z-3)/(-2)

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