Write down a unit vector in XY-plane, making an angle of 30 degrees with the positive direction of x-axis.
Hint. A unit vector at angle theta from the x-axis in the XY-plane has components (cos(theta), sin(theta), 0).
Since a unit vector at angle theta from the x-axis has components (cos(theta), sin(theta), 0), substituting theta=30 degrees gives cos(30)=sqrt3/2 and sin(30)=1/2.
✦ (sqrt(3)/2)i+(1/2)j
Find the scalar components and magnitude of the vector joining the points P(x1,y1,z1) and Q(x2,y2,z2).
Hint. Subtract the coordinates of P from Q to get the components, then apply the magnitude formula.
Since the vector PQ equals Q minus P, its scalar components are (x2-x1), (y2-y1), (z2-z1), and its magnitude is the square root of the sum of the squares of these three components.
✦ Scalar components: (x2-x1), (y2-y1), (z2-z1). Magnitude: sqrt[(x2-x1)^2+(y2-y1)^2+(z2-z1)^2]
A girl walks 4 km towards west, then she walks 3 km in a direction 30 degrees east of north and stops. Determine the girl's displacement from her initial point of departure.
Hint. Represent each leg of the walk as a vector, then add them to find the net displacement.
Taking east and north as the positive x and y directions, the first leg is (-4,0) and the second leg is 3(sin30,cos30)=(1.5, 1.5sqrt3); adding these gives the net displacement vector, whose magnitude is sqrt13.
✦ Net displacement: (-2.5)i+(1.5sqrt(3))j km, with magnitude sqrt(13) km
If a=b+c, then is it true that |a|=|b|+|c|? Justify your answer.
Hint. Consider the triangle inequality, and check whether it becomes an equality only in a special case.
This is not generally true, since the triangle inequality only gives |b+c| less than or equal to |b|+|c|; equality holds only when b and c point in exactly the same direction, which means |a| is strictly less than |b|+|c| whenever they point in different directions.
✦ Not necessarily true — equality holds only when b and c are in the same direction
Find the value of x for which x(i+j+k) is a unit vector.
Hint. Set the magnitude of x(i+j+k) equal to 1 and solve for x.
Since |x(i+j+k)|=|x|.sqrt3, setting this equal to 1 gives |x|=1/sqrt3, so x can be either sign.
✦ x=+-1/sqrt(3)
Find a vector of magnitude 5 units, and parallel to the resultant of the vectors a=2i+3j-k and b=i-2j+k.
Hint. First find the resultant a+b, then scale it to magnitude 5.
Since the resultant is a+b=3i+j+0k, with magnitude sqrt10, scaling it by 5/sqrt10 gives a vector in the same direction with magnitude exactly 5.
✦ (15/sqrt(10))i+(5/sqrt(10))j
If a=i+j+k, b=2i-j+3k and c=i-2j+k, find a unit vector parallel to the vector 2a-b+3c.
Hint. Compute the combination 2a-b+3c component by component, then normalise the result.
Computing 2a-b+3c gives (2-2+3, 2+1-6, 2-3+3)=(3,-3,2), with magnitude sqrt22, so dividing by sqrt22 gives the unit vector.
✦ (3/sqrt(22))i-(3/sqrt(22))j+(2/sqrt(22))k
Show that the points A(1,-2,-8), B(5,0,-2) and C(11,3,7) are collinear, and find the ratio in which B divides AC.
Hint. Form vectors AB and AC and check whether one is a scalar multiple of the other, then compare their magnitudes for the ratio.
Since AB=(4,2,6) and AC=(10,5,15)=2.5.AB, the three points are collinear; because AB is 2/5 of AC, B divides AC in the ratio 2:3.
✦ Collinear; B divides AC in the ratio 2:3
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2a+b) and (a-3b) externally in the ratio 1:2. Also, show that P is the mid point of the line segment RQ.
Hint. Apply the external section formula with m=1, n=2, then verify the midpoint claim directly.
Using the external division formula R=(m.Q-n.P)/(m-n) with m=1, n=2 gives R=[(a-3b)-2(2a+b)]/(-1)=3a+5b; averaging R and Q then gives (R+Q)/2=(3a+5b+a-3b)/2=2a+b, which is exactly P, confirming the midpoint claim.
✦ R=3a+5b; and (R+Q)/2=2a+b=P, confirming P is the midpoint of RQ
The two adjacent sides of a parallelogram are 2i-4j+5k and i-2j-3k. Find the unit vector parallel to its diagonal. Also, find its area.
Hint. The diagonal from the common vertex is the sum of the two adjacent sides; the area is the magnitude of their cross product.
Since the diagonal is the sum (3,-6,2) with magnitude 7, dividing through gives the unit vector; because the area equals the magnitude of the cross product of the two sides, computing that determinant gives 11.sqrt5.
✦ Unit vector: (3/7)i-(6/7)j+(2/7)k. Area: 11.sqrt(5) square units
Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are +-(1/sqrt3, 1/sqrt3, 1/sqrt3).
Hint. Let the vector make equal angles with all three axes, so all three direction cosines are equal, then use l^2+m^2+n^2=1.
If the vector is equally inclined to all three axes, its direction cosines l, m, n are all equal to some common value t; since l^2+m^2+n^2=1 for any set of direction cosines, 3t^2=1, giving t=+-1/sqrt3 for all three simultaneously.
✦ Proved: direction cosines are +-(1/sqrt(3), 1/sqrt(3), 1/sqrt(3))
Let a=i+4j+2k, b=3i-2j+7k and c=2i-j+4k. Find a vector d which is perpendicular to both a and b, and c.d=15.
Hint. Find a vector n perpendicular to both a and b via their cross product, then scale it so that c.d=15.
The cross product of a and b gives n=(32,-1,-14), perpendicular to both; since c.n=9, scaling n by 15/9=5/3 gives a vector d=(5/3)n that is still perpendicular to a and b while satisfying c.d=15.
✦ d=(160/3)i-(5/3)j-(70/3)k
The scalar product of the vector i+j+k with a unit vector along the sum of vectors 2i+4j-5k and lambda.i+2j+3k is equal to one. Find the value of lambda.
Hint. Add the two vectors (keeping lambda symbolic), form the unit vector along the sum, then set the dot product with i+j+k equal to 1.
The sum is (2+lambda, 6, -2), so the equation (1)(2+lambda)+(1)(6)+(1)(-2) equal to the magnitude sqrt[(2+lambda)^2+36+4] simplifies (after squaring both sides) to a linear equation in lambda.
✦ lambda=1
If a, b, c are mutually perpendicular vectors of equal magnitude, show that the vector a+b+c is equally inclined to a, b and c. [Note: the book's own printed statement of this question contains a leftover reference to "c.d=15" carried over from the previous question, which does not belong to this one — the intended and standard question, solved here, is the equal-inclination result stated above.]
Hint. Compute the angle between (a+b+c) and each of a, b, c individually, using that a,b,c are mutually perpendicular (so cross dot products vanish) and of equal magnitude.
Since a, b, c are mutually perpendicular, a.b=b.c=c.a=0, so (a+b+c).a=|a|^2 (and similarly for b and c); with |a|=|b|=|c|=k, each of these dot products equals k^2, and each individual magnitude |a|,|b|,|c| is the same k, so the cosine of the angle between (a+b+c) and each of a, b, c works out to the same value in every case.
✦ Proved: a+b+c makes the same angle with each of a, b, c, since (a+b+c).a=(a+b+c).b=(a+b+c).c=k^2 by mutual perpendicularity
Prove that (a+b).(a+b)=|a|^2+|b|^2, if and only if a and b are perpendicular, given a is not 0, b is not 0.
Hint. Expand the dot product on the left using distributivity, then compare to the right side.
Expanding gives (a+b).(a+b)=|a|^2+2(a.b)+|b|^2; this equals |a|^2+|b|^2 exactly when 2(a.b)=0, i.e. when a.b=0, which (since neither vector is zero) happens exactly when a and b are perpendicular.
✦ Proved: the identity holds if and only if a.b=0, i.e. a is perpendicular to b
Choose the correct answer: if theta is the angle between two vectors a and b, then a.b is greater than or equal to 0 only when (A) 0<theta<pi/2 (B) 0<=theta<=pi/2 (C) 0<theta<pi (D) 0<=theta<=pi
Hint. Recall that a.b=|a||b|cos(theta), and cos(theta) is non-negative exactly on a specific range of theta.
Since a.b=|a||b|cos(theta) and magnitudes are always non-negative, a.b>=0 exactly when cos(theta)>=0, which holds for theta in the closed range from 0 to pi/2 inclusive.
✦ (B) 0<=theta<=pi/2
Choose the correct answer: let a and b be two unit vectors and theta is the angle between them. Then a+b is a unit vector if theta is (A) pi/4 (B) pi/3 (C) pi/2 (D) 2pi/3
Hint. Set |a+b|^2=|a|^2+2(a.b)+|b|^2 equal to 1 and solve for theta, using |a|=|b|=1.
Since |a+b|^2=1+2cos(theta)+1=2+2cos(theta), setting this equal to 1^2=1 gives cos(theta)=-1/2, so theta=2pi/3.
✦ (D) theta=2pi/3
Choose the correct answer: the value of i.(j x k)+j.(i x k)+k.(i x j) is (A) 0 (B) -1 (C) 1 (D) 3
Hint. Recall the standard identities j x k=i, i x k=-j, i x j=k, then substitute and simplify.
Substituting the standard cross products gives i.i+j.(-j)+k.k=1-1+1=1, since i.i=k.k=1 and j.j=1 (so -j.j=-1).
✦ (C) 1
Choose the correct answer: if theta is the angle between any two vectors a and b, then |a.b|=|a x b| when theta is equal to (A) 0 (B) pi/4 (C) pi/2 (D) pi
Hint. Set |a||b|cos(theta) equal to |a||b|sin(theta) and solve for theta.
Since |a.b|=|a||b|.|cos(theta)| and |a x b|=|a||b|.sin(theta) (for theta in [0,pi] where sin is non-negative), setting |cos(theta)|=sin(theta) gives tan(theta)=1 (taking the case cos>0), so theta=pi/4.
✦ (B) pi/4
Solutions written by the tuition.in editorial team and checked against lemh204.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.
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