Vector Algebra — Practice

Question 1

Classify time period, distance, and force as scalar or vector quantities.

Accepted Answer

Time period and distance are fully described by a single number; force has both a magnitude and a direction. Answer: Time period and distance are scalars; force is a vector.

Question 2

Find the unit vector in the direction of a=i+j+2k.

Accepted Answer

The magnitude is sqrt(1+1+4)=sqrt6, so dividing each component by sqrt6 gives the unit vector. Answer: (1/sqrt6)i+(1/sqrt6)j+(2/sqrt6)k.

Question 3

Find the position vector of the midpoint of the vector joining P(2,3,4) and Q(4,1,-2).

Accepted Answer

The midpoint is the average of the two position vectors: (2+4)/2, (3+1)/2, (4-2)/2. Answer: 3i+2j+k.

Question 4

Find the projection of the vector i-j on the vector i+j.

Accepted Answer

Since (i-j).(i+j)=1-1=0, the projection is 0 divided by the magnitude of i+j. Answer: 0 (the two vectors are perpendicular).

Question 5

Find the area of the triangle with vertices A(1,1,2), B(2,3,5), C(1,5,5).

Accepted Answer

With AB=(1,2,3) and AC=(0,4,3), the cross product is (-6,-3,4), with magnitude sqrt61; the triangle's area is half of that. Answer: sqrt(61)/2 square units.

Question 6

If a, b, c are unit vectors such that a+b+c=0, find a.b+b.c+c.a.

Accepted Answer

Squaring a+b+c=0 gives |a|^2+|b|^2+|c|^2+2(a.b+b.c+c.a)=0, and since each vector is a unit vector, 3+2(a.b+b.c+c.a)=0. Answer: a.b+b.c+c.a=-3/2.

Question 7

Show that (a-b) x (a+b)=2(a x b).

Accepted Answer

Expanding gives a x a+a x b-b x a-b x b, and since a x a=b x b=0 and b x a=-(a x b), this simplifies to a x b+a x b. Answer: Proved: (a-b) x (a+b)=2(a x b).

Question 8

Find the position vector of R which divides PQ (position vectors i+2j-k and -i+j+k) externally in the ratio 2:1.

Accepted Answer

Using R=(m.q-n.p)/(m-n) with m=2, n=1 gives R=2(-i+j+k)-1(i+2j-k), divided by 1. Answer: R=-3i+3k.

Practice — Vector Algebra

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