Vector Algebra
Distance from point to line
Grade 12
Question:
<p>The co-ordinates of the point P on the line \[\mathbf{r} = (\mathbf{i} + \mathbf{j} + \mathbf{k}) + \lambda(-\mathbf{i} + \mathbf{j} - \mathbf{k})\] which is nearest to the origin is</p>
<p>(a) \(\left(\frac{2}{3}, \frac{4}{3}, \frac{2}{3}\right)\)</p>
<p>(b) \(\left(-\frac{2}{3}, -\frac{4}{3}, \frac{2}{3}\right)\)</p>
<p>(c) \(\left(\frac{2}{3}, \frac{4}{3}, -\frac{2}{3}\right)\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The point on a line nearest to the origin lies where the position vector from origin to that point is perpendicular to the direction vector of the line. Use the perpendicularity condition (dot product = 0) to find the parameter λ.
Step 1: Write the parametric form of the line. The line is given as: r = ( i + j + k ) + λ(- i + j - k ) A general point P on the line has position vector: r (λ) = (1-λ, 1+λ, 1-λ) Step 2: Identify the direction vector and apply the perpendicularity condition. The direction vector of the line is d = (-1, 1, -1) For the nearest point to origin O, vector OP must be perpendicular to d r (λ) · d = 0 Step 3: Calculate the dot product. (1-λ, 1+λ, 1-λ) · (-1, 1, -1) = 0 -1(1-λ) + 1(1+λ) + (-1)(1-λ) = 0 -(1-λ) + (1+λ) - (1-λ) = 0 -1 + λ + 1 + λ - 1 + λ = 0 3λ - 1 = 0 λ = 1/3 Step 4: Find the coordinates of P. Substituting λ = 1/3 into the parametric equations: x = 1 - (1/3) = 2/3 y = 1 + (1/3) = 4/3 z = 1 - (1/3) = 2/3 Therefore, P = (2/3, 4/3, 2/3) Step 5: Verify against options. This matches option (a): (2/3, 4/3, 2/3) ∴ Answer: Unknown
Correct Answer: Unknown