GATE 2023 EC Q.11: solution and answer

GATE 2023 · ECQ.11 · 1 mark · MCQ

Let $v_{1} = 120$ and $v_{2} = 213$ be two vectors. The value of the coefficient $α$ in the expression $v_{1} = α v_{2} + e$ , which minimizes the length of the error vector $e$ , is

(A)
$\frac{7}{2}$
(B)
$\frac{- 2}{7}$
(C)
$\frac{2}{7}$
(D)
$\frac{- 7}{2}$

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Answer and solution

ANSWER

$α = \frac{v _{1} \cdot v _{2}}{∥ v _{2} ∥ ^{2}} = \frac{4}{14} = \frac{2}{7}$ , option (C).

Official answer: CSource: Official answer key, GATE 2023 EC (IIT Kanpur)

The error vector is $e = v_{1} - α v_{2}$ . Its length is smallest when $α v_{2}$ is the orthogonal projection of $v_{1}$ onto $v_{2}$ , i.e. when $e ⊥ v_{2}$ .
Orthogonality condition: $v_{2}^{T} (v_{1} - α v_{2}) = 0 \Rightarrow α = \frac{v _{1} \cdot v _{2}}{v _{2} \cdot v _{2}}$ .
The same result follows from calculus: $∥ e ∥^{2} = ∥ v_{1} ∥^{2} - 2 α (v_{1} \cdot v_{2}) + α^{2} ∥ v_{2} ∥^{2}$ . Setting $\frac{d}{d α} ∥ e ∥^{2} = - 2 (v_{1} \cdot v_{2}) + 2 α ∥ v_{2} ∥^{2} = 0$ gives the same $α$ , and the second derivative $2∥ v_{2} ∥^{2} > 0$ confirms it is a minimum.
Dot product: $v_{1} \cdot v_{2} = (1) (2) + (2) (1) + (0) (3) = 4$ .
Squared norm: $v_{2} \cdot v_{2} = 2^{2} + 1^{2} + 3^{2} = 14$ .
Therefore $α = \frac{4}{14} = \frac{2}{7}$ , which is option (C).
Check: $e = 120 - \frac{2}{7} 213 = \frac{1}{7} 312 - 6$ , and $e \cdot v_{2} = \frac{1}{7} (6 + 12 - 18) = 0$ , so $e$ is orthogonal to $v_{2}$ as required.

KEY IDEA

The error is shortest when it is perpendicular to $v_{2}$ , which makes $α v_{2}$ the orthogonal projection of $v_{1}$ onto $v_{2}$ . So $α = \frac{v _{1} \cdot v _{2}}{v _{2} \cdot v _{2}}$ (least squares with a single unknown).

WATCH OUT FOR

Swapping the roles of the vectors and dividing by $∥ v_{1} ∥^{2} = 5$ (which gives $4/5$ ), or inverting the ratio to get $7/2$ (option A). The denominator is the squared length of the vector you are projecting onto, here $v_{2}$ , so it is $14$ . A sign slip in the dot product leads to the negative options B or D.

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GATE 2023 EC Q.11: solution and answer

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