Vector Algebra — Practice

Question 1

The unit vector in the direction of a = 3i − 4j + 12k is:
(a) (3/13)i − (4/13)j + (12/13)k  (b) (3/5)i − (4/5)j + (12/5)k  (c) 3i − 4j + 12k  (d) (3/19)i − (4/19)j + (12/19)k

Accepted Answer

|a| = √(3² + (−4)² + 12²) = √(9+16+144) = √169 = 13. Unit vector = a/|a| = (3/13)i − (4/13)j + (12/13)k.

Answer: (a). Trap (b) uses magnitude 5 = √(9+16), computed from only the i and j components and forgetting the k-component entirely under the square root. Trap (c) is the vector itself with no normalisation at all — a very literal 'forgot to divide' error. Trap (d) divides by 19 = 3+4+12, the sum of the (absolute) components, instead of √(sum of squares) — confusing magnitude with a simple component sum.

Question 2

If a vector a = i + 2j + 2k makes direction cosines (l, m, n) with the coordinate axes, then (l, m, n) equals:
(a) (1/3, 2/3, 2/3)  (b) (1, 2, 2)  (c) (1/5, 2/5, 2/5)  (d) (1/9, 2/9, 2/9)

Accepted Answer

|a| = √(1²+2²+2²) = √9 = 3. Direction cosines l=1/|a|=1/3, m=2/3, n=2/3. Check: (1/3)²+(2/3)²+(2/3)² = 1/9+4/9+4/9 = 1 ✓.

Answer: (a). Trap (b) gives the direction RATIOS (the raw components), not direction cosines — they don't satisfy l²+m²+n²=1 (1²+2²+2²=9≠1). Trap (c) divides by 5 = 1+2+2, the sum of the components, instead of the magnitude 3. Trap (d) divides by 9 = |a|², using the squared magnitude instead of the magnitude itself.

Question 3

Two vectors of magnitudes 6 and 10 act at a point with an angle of 60° between them. The magnitude of their resultant is:
(a) 14  (b) 16  (c) √136  (d) 4

Accepted Answer

|a+b| = √(|a|²+|b|²+2|a||b|cosθ) = √(36+100+2×6×10×cos60°) = √(136+120×0.5) = √(136+60) = √196 = 14.

Answer: (a). Trap (b) 16 simply adds the two magnitudes (6+10), as if the vectors were collinear at θ=0°. Trap (c) √136 = √(6²+10²) drops the cross term 2|a||b|cosθ entirely, treating vector addition like plain Pythagoras regardless of the angle. Trap (d) 4 = 10−6 = |a−b|, using θ=180° (vectors opposing) instead of the given 60°.

Question 4

The position vectors of the vertices of triangle ABC are a = 2i+3j+4k, b = 4i−j+2k and c = 4j+6k. The position vector of the centroid of the triangle is:
(a) 2i+2j+4k  (b) 6i+6j+12k  (c) 3i+3j+6k  (d) 3i+j+3k

Accepted Answer

Centroid = (a+b+c)/3 = [(2+4+0)i + (3−1+4)j + (4+2+6)k]/3 = (6i+6j+12k)/3 = 2i+2j+4k.

Answer: (a). Trap (b) is the raw sum a+b+c with the ÷3 forgotten entirely. Trap (c) divides the sum by 2 instead of 3 — confusing the centroid formula (÷3, three vertices) with the midpoint formula (÷2, two points). Trap (d) computes (a+b)/2, the midpoint of just two of the three vertices, ignoring c altogether.

Question 5

The position vectors of points A and B are a = i+2j+3k and b = 6i−3j+8k. The position vector of the point P that divides AB internally in the ratio 2:3 is:
(a) 3i+5k  (b) 15i+25k  (c) 4i−j+6k  (d) −9i+12j−7k

Accepted Answer

For internal division in ratio m:n = 2:3, r = (m·b+n·a)/(m+n) = (2b+3a)/5.

2b = 12i−6j+16k; 3a = 3i+6j+9k; sum = 15i+0j+25k. Divide by 5: r = 3i+5k.

Answer: (a). Trap (b) 15i+25k is exactly the numerator (2b+3a) with the ÷(m+n)=÷5 step forgotten. Trap (c) swaps which vertex gets which weight — computing (3b+2a)/5 instead of (2b+3a)/5, i.e. applying the ratio in reverse order. Trap (d) −9i+12j−7k comes from mistakenly using the EXTERNAL division formula (2b−3a)/(2−3) instead of the internal one — the question explicitly says 'internally'.

Question 6

For what value of λ are the vectors a = 2i + λj + k and b = i − 2j + 3k perpendicular to each other?
(a) 5/2  (b) −5/2  (c) 5  (d) 2/5

Accepted Answer

a·b = 0 for perpendicular vectors: (2)(1) + λ(−2) + (1)(3) = 0 ⟹ 2 − 2λ + 3 = 0 ⟹ 5 − 2λ = 0 ⟹ λ = 5/2.

Answer: (a). Trap (b) is a sign error, flipping the final answer's sign. Trap (c) comes from mishandling the coefficient on λ — solving 5−λ=0 instead of 5−2λ=0, i.e. treating the −2λ term as if it were just −λ. Trap (d) inverts the correct fraction (2/5 instead of 5/2), a common careless slip when writing the final answer under time pressure.

Question 7

The cosine of the angle between the vectors a = i+2j+2k and b = 2i+2j+k is:
(a) 8/9  (b) 8/25  (c) 2/3  (d) 4/3

Accepted Answer

a·b = (1)(2)+(2)(2)+(2)(1) = 2+4+2 = 8. |a| = √(1+4+4) = 3, |b| = √(4+4+1) = 3. cosθ = (a·b)/(|a||b|) = 8/(3×3) = 8/9.

Answer: (a). Trap (b) 8/25 uses the SUM of the components (1+2+2=5 and 2+2+1=5) as if that were the magnitude, instead of √(sum of squares) — confusing magnitude with a component sum. Trap (c) 2/3 drops the third term of the dot product, computing only (1)(2)+(2)(2)=6 and giving 6/9=2/3. Trap (d) 4/3 adds the two magnitudes (3+3=6) instead of multiplying them (3×3=9), giving 8/6=4/3 — a value greater than 1, which is impossible for a cosine, but a natural output of the 'add instead of multiply' slip.

Question 8

The scalar projection of a = 3i − j + 2k on b = i + 2j − 2k is:
(a) −1  (b) 1  (c) −3  (d) −1/3

Accepted Answer

a·b = (3)(1)+(−1)(2)+(2)(−2) = 3−2−4 = −3. |b| = √(1+4+4) = 3. Scalar projection = (a·b)/|b| = −3/3 = −1.

Answer: (a). Trap (b) 1 drops the sign, forgetting that a negative scalar projection is a perfectly valid (and here, correct) result when the angle between the vectors is obtuse. Trap (c) −3 is just a·b with the ÷|b| step skipped entirely. Trap (d) −1/3 divides by |b|²=9 instead of |b|=3 — this is the exact projection slip the chapter flags most often: the vector-projection coefficient uses |b|², but the SCALAR projection asked for here uses |b| alone.

Question 9

The position vectors of the vertices of a triangle are a = i+j+k, b = 2i+3j+k and c = 3i+j+2k. The area of the triangle is:
(a) √21/2  (b) √21  (c) 1  (d) 5√3/2

Accepted Answer

AB = b−a = i+2j, AC = c−a = 2i+k. AB×AC = |i j k; 1 2 0; 2 0 1| = (2×1−0×0)i − (1×1−0×2)j + (1×0−2×2)k = 2i−j−4k. |AB×AC| = √(4+1+16) = √21. Area of triangle = ½√21 = √21/2.

Answer: (a). Trap (b) √21 forgets the ÷2 that converts the parallelogram's area (|AB×AC|) into the triangle's area. Trap (c) 1 confuses cross and dot product, computing ½|AB·AC| = ½(1×2+2×0+0×1) = ½(2) = 1 instead of using the cross product at all. Trap (d) 5√3/2 comes from forgetting to subtract the common vertex A before taking the cross product — computing ½|b×c| directly from the raw position vectors instead of ½|AB×AC|, which gives a completely different, geometrically meaningless pair of 'side vectors'.

Question 10

A unit vector perpendicular to both a = i+j+k and b = 2i−j+2k is:
(a) (1/√2)i − (1/√2)k  (b) −(1/√2)i + (1/√2)k  (c) i − k  (d) (1/3)i − (1/3)k

Accepted Answer

a×b = |i j k; 1 1 1; 2 −1 2| = (1×2−1×(−1))i − (1×2−1×2)j + (1×(−1)−1×2)k = (2+1)i − (2−2)j + (−1−2)k = 3i+0j−3k. |a×b| = √(9+0+9) = √18 = 3√2. Unit vector = (3i−3k)/(3√2) = (i−k)/√2 = (1/√2)i − (1/√2)k.

Answer: (a). Trap (b) is the unit vector along b×a instead of a×b — since the cross product is anti-commutative, reversing the order flips every component's sign, giving −(1/√2)i + (1/√2)k. Trap (c) i−k is the cross product with the common factor of 3 cancelled out, but never actually divided by the magnitude — it isn't a unit vector at all (its own magnitude is √2, not 1). Trap (d) uses |a×b| = 3 instead of the correct 3√2, dropping the √2 factor when taking the square root of 18.

Question 11

The volume of the parallelepiped whose coterminous edges are a = 2i−j+k, b = i+2j−k and c = i−j+2k is:
(a) 6  (b) −6  (c) 0  (d) 9

Accepted Answer

Volume = |[a b c]| = |a·(b×c)|, computed directly as the determinant |2 −1 1; 1 2 −1; 1 −1 2| = 2(2×2−(−1)×(−1)) − (−1)(1×2−(−1)×1) + 1(1×(−1)−2×1) = 2(4−1) + 1(2+1) + 1(−1−2) = 2(3)+1(3)+1(−3) = 6+3−3 = 6.

Answer: (a). Trap (b) −6 comes from evaluating [b a c] instead of [a b c] — swapping two of the three vectors flips the sign of the scalar triple product, so a mixed-up row order gives the right magnitude but the wrong sign. Trap (c) 0 wrongly assumes the vectors are coplanar without actually evaluating the determinant. Trap (d) 9 comes from dropping the third cofactor term of the row-1 expansion entirely — computing only 2(3) − (−1)(3) = 6+3 = 9 and forgetting to add the final 1×(−3) = −3 term.

Question 12

The vectors a = 2i − 4j + 6k and b = i − 2j + mk are parallel for m equal to:
(a) 3  (b) 6  (c) −3  (d) 12

Accepted Answer

For parallel vectors, corresponding components are proportional: a1/b1 = a2/b2 = a3/b3. Here 2/1 = −4/−2 = 2, so 6/m must also equal 2, giving m = 3. (Check: a = 2b requires b = i−2j+3k, which matches when m=3.)

Answer: (a). Trap (b) 6 just copies a's k-component directly as m without applying the common ratio of 2. Trap (c) −3 is a sign error, using a = −2b instead of a = 2b. Trap (d) 12 multiplies by the ratio (6×2) instead of dividing by it (6÷2), inverting the proportionality direction.

Practice — Vector Algebra

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