Q13

1 markMCQSection A

The value of p for which the vectors $\hat{i} + 2\hat{j} + 3\hat{k}$ and $2\hat{i} - p\hat{j} + \hat{k}$ are perpendicular to each other is

Q: The value of p for which the vectors $\hat{i} + 2\hat{j} + 3\hat{k}$ and $2\hat{i} - p\hat{j} + \hat{k}$ are perpendicular to each other is

**Correct option: (C) $\frac{5}{2}$** Two vectors are perpendicular when their dot product is zero. Setting $(i+2j+3k) \cdot (2i-pj+k) = 2 - 2p + 3 = 5 - 2p = 0$ gives $p = \frac{5}{2}$.

Q: What is the condition for two vectors to be perpendicular?

Their dot (scalar) product must equal zero: $a \cdot b = 0$, provided neither vector is the zero vector.

Vector Algebra

Perpendicular Vectors — Dot Product Condition

Options

(A)0

(B)1

(C)

\frac{5}{2}

(D)

-\frac{5}{2}

Official Answer

Correct option: (C) $\frac{5}{2}$

Two vectors are perpendicular when their dot product is zero. Setting $(i+2j+3k) \cdot (2i-pj+k) = 2 - 2p + 3 = 5 - 2p = 0$ gives $p = \frac{5}{2}$ .

dot productperpendicular vectorsscalar productvector algebracondition for perpendicularityzero dot product

Marking Scheme

11 mark: awarded only for selecting option (C) $\frac{5}{2}$ ; no partial credit.

Hint

Set the dot product of the two vectors equal to zero and solve the resulting linear equation for p.

Quick Oral Answer

Perpendicular vectors have zero dot product; computing $(1)(2)+(2)(-p)+(3)(1)=0$ gives $5-2p=0$ , so $p=\frac{5}{2}$ .

Analysis & Explanation

Concept: Dot product = 0 is the necessary and sufficient condition for perpendicularity of two non-zero vectors.

For $a = i+2j+3k$ and $b = 2i-pj+k$ , $a \cdot b = (1)(2) + (2)(-p) + (3)(1) = 2 - 2p + 3 = 5 - 2p$ . Perpendicularity requires $a \cdot b = 0$ , so $5 - 2p = 0$ , giving $p = \frac{5}{2}$ .

Why the distractors fail:

(A) 0 would make a.b = 5, not 0, so the vectors would not be perpendicular; this likely comes from forgetting to solve the equation and picking a 'default' value.
(B) 1 gives a.b = 5 - 2 = 3, not 0, an arithmetic slip in solving $5 - 2p = 0$ .
(D) -5/2 is the correct magnitude with the wrong sign, typically from an error transposing terms across the equation (writing 5 + 2p = 0 instead of $5 - 2p = 0$ ).

Exam tip: Carefully track the sign of each component when computing the dot product component-wise — a single sign slip changes the final answer completely.

Common Mistakes

1Sign errors while multiplying components, especially with the -p term.
2Forgetting that perpendicularity requires the dot product to be exactly 0, not merely comparing magnitudes.
3Transposition errors leading to the wrong-signed answer ( $-\frac{5}{2}$ instead of $\frac{5}{2}$ ).

Interesting Facts

The dot product condition $a \cdot b = 0$ for perpendicularity is a direct 3D generalization of the 2D condition that the product of slopes of two perpendicular lines is -1.

Vector perpendicularity checks via the dot product are used extensively in computer graphics to determine surface normals and lighting angles.

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Frequently Asked Questions

What is the condition for two vectors to be perpendicular?

Their dot (scalar) product must equal zero: $a \cdot b = 0$ , provided neither vector is the zero vector.

Could there be more than one value of p satisfying this?

No — since the dot product here is a linear expression in p ( $5-2p$ ), setting it to zero gives exactly one solution, $p = \frac{5}{2}$ .