Question 232 marksVery Short AnswerSection B

Find the vector of magnitude 14 in the direction of vector QP, where P and Q are the points (1,3,2)(1, 3, 2) and (−1,0,8)(-1, 0, 8) respectively.

Vector Algebra
Vectors - Unit Vector and Scalar Multiple

Official answer

The required vector is 4i+6j−12k4i + 6j - 12k (magnitude 14).


Working: Vector QP = position of P − position of Q = (1−(−1),3−0,2−8)=(2,3,−6)(1-(-1), 3-0, 2-8) = (2, 3, -6), with ∣QP∣=4+9+36=7|QP| = \sqrt{4+9+36} = 7. The unit vector along QP is (2/7, 3/7, −6/7), so the vector of magnitude 14 in this direction is 14×(2/7,3/7,−6/7)=(4,6,−12)14 \times (2/7, 3/7, -6/7) = (4, 6, -12), i.e., 4i+6j−12k4i + 6j - 12k.

Answer keywords:vector QPunit vectormagnitude of a vectorposition vectordirection of a vectorscalar multiplication of vectors

Marking scheme

  1. 1.1 mark: correctly finding QP=2i+3j−6kQP = 2i+3j-6k and its magnitude 7.
  2. 2.1 mark: correctly scaling the unit vector by 14 to get 4i+6j−12k4i+6j-12k.

Hint

QP=(position vector of P)−(position vector of Q)QP = (\text{position vector of } P) - (\text{position vector of } Q); find its magnitude, then scale the unit vector by 14.

Quick oral answer

I find QP=P−Q=(2,3,−6)QP = P - Q = (2,3,-6), its magnitude 7, then multiply the unit vector (2/7,3/7,−6/7)(2/7,3/7,-6/7) by 14 to get (4,6,−12)(4,6,-12).

Common mistakes

  1. 1.Computing PQ (P to Q) instead of QP (Q to P), reversing the sign of every component.
  2. 2.Forgetting to divide by the magnitude before multiplying by 14, giving a vector of the wrong length.
  3. 3.Arithmetic slip in computing the magnitude 4+9+36=49=7\sqrt{4+9+36} = \sqrt{49} = 7.

Analysis and explanation

This question tests finding a vector of a specified magnitude along a given direction — a two-step process: find the direction vector, then scale it.


Concept: Any vector of magnitude m along vector v is given by m⋅(v/∣v∣)m \cdot (v/|v|), where v/∣v∣v/|v| is the unit vector in the direction of v.


Key step — order of points: QP means the vector FROM Q TO P, i.e., (position of P) − (position of Q), not the reverse. Reversing the order gives PQ = −QP, which points in the opposite direction and would be marked wrong.


Exam trap: Many students compute PQ (from P to Q) instead of QP by mistake, or forget to first find the unit vector before scaling to magnitude 14.


Real-world link: Scaling a direction vector to a specified magnitude is exactly how force vectors, velocity vectors, or displacement vectors of a known size (e.g., a 14 N force in a certain direction) are constructed in physics and engineering.

Interesting facts

Since |QP| = 7 divides evenly into the required magnitude 14, this problem is designed so the unit vector scales by a clean factor of 2 — a common CBSE design choice to keep numbers simple.

The same unit-vector scaling technique is used in computer graphics and robotics to set an object's velocity or force to an exact magnitude while preserving a given direction.

Question details

Q23 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Vector Algebra
Topic
Vectors - Unit Vector and Scalar Multiple
Bloom's level
apply
Syllabus topic
Unit 4: Vector Algebra — Direction of a Vector, Vector of Given Magnitude
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

What does the vector QP mean geometrically?

QP denotes the vector starting at point Q and ending at point P, computed as (position vector of P)−(position vector of Q)(\text{position vector of } P) - (\text{position vector of } Q).

How do you construct a vector of a specific magnitude in a given direction?

Divide the given direction vector by its own magnitude to get a unit vector, then multiply that unit vector by the desired magnitude.