Question

If the equation of the straight line passing through the point \((a,1,3)\) and parallel to the vector \(\frac{2}{3}\hat{i} + \frac{3}{2}\hat{j} + \hat{k}\) is \(\frac{3x+6}{b} = \frac{2y-2}{3} = \frac{z-3}{1}\), then the value of \(a+b\) is equal to

• A. 3
• B. -3
• C. -1
• D. 1
• E. 0

Hint:

Before comparing denominators in 3D line equations, always ensure that the coefficients of \(x, y,\) and \(z\) in the numerators are exactly \(+1\). This is the most common place where errors occur.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The standard equation of a line passing through \((x_1, y_1, z_1)\) and parallel to vector \((l, m, n)\) is: \[ \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \]

Step 2: Key Formula or Approach:
Standardize the given equation \(\frac{3x+6}{b} = \frac{2y-2}{3} = \frac{z-3}{1}\) by making the coefficients of \(x, y, z\) in the numerators equal to 1. Then compare with the given point and parallel vector.

Step 3: Detailed Explanation:
1. Standardize the given equation: \[ \frac{3(x+2)}{b} = \frac{2(y-1)}{3} = \frac{z-3}{1} \] \[ \frac{x+2}{b/3} = \frac{y-1}{3/2} = \frac{z-3}{1} \]
2. Comparison with point \((a, 1, 3)\): The numerator of the \(x\)-term is \(x - (-2)\), so the point is \((-2, 1, 3)\). Thus, \(a = -2\).
3. Comparison with parallel vector \(\frac{2}{3}\hat{i} + \frac{3}{2}\hat{j} + \hat{k}\): The denominators must be proportional to the vector components \((2/3, 3/2, 1)\). The \(z\)-denominator is 1, matching the vector component exactly. So, the \(x\)-denominator must be: \[ \frac{b}{3} = \frac{2}{3} \implies b = 2 \]
4. Final calculation: \[ a + b = -2 + 2 = 0 \]

Step 4: Final Answer:
The value of \(a+b\) is 0.