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Question #40682

Difficulty: 8/10 | 🟢 Free

The Question

Let $x, y, z$ be three consecutive terms of an arithmetic progression with common difference $d$ . Let $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ be a vector such that its $y$ -component is $2$ . Consider a variable vector $\vec{n} = (d+1)\hat{i} + \hat{j} + \hat{k}$ . If $M$ is the maximum possible value of the scalar projection of $\vec{v}$ on $\vec{n}$ for $d \in \mathbb{R}$ , then $M$ equals:

Options

A.
$2\sqrt{3}$
B.
$7$
C.
$\frac{7}{\sqrt{6}}$
D.
$6$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Selection of AP Terms|Dot Product Projection Sign|Unit Vector Normalization

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between the dot product and the scalar projection, specifically when the normal vector is variable.

A common trap is to maximize the unnormalized dot product $\vec{v} \cdot \vec{n} = -d^2 + 2d + 6$ , which yields a maximum at $d=1$ .

Students falling for this will either select the maximum dot product value of 7, or incorrectly substitute $d=1$ into the projection formula to get $7/\sqrt{6}$ .

The correct approach requires normalizing the variable vector $\vec{n}$ before differentiating, which shifts the critical point to $d=0$ and yields the true maximum projection of $2\sqrt{3}$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	Selection of AP Terms
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free