Find the angle between two vectors a and b with magnitudes sqrt3 and 2, respectively having a.b=sqrt6.
Hint. Use the formula a.b=|a||b|cos(theta) and solve for theta.
Since a.b=|a||b|cos(theta), substituting gives sqrt6=sqrt3.2.cos(theta), so cos(theta)=sqrt6/(2sqrt3)=1/sqrt2, giving theta=pi/4.
✦ theta=pi/4 (45 degrees)
Find the angle between the vectors i-2j+3k and 3i-2j+k.
Hint. Compute the dot product and both magnitudes, then apply cos(theta)=a.b/(|a||b|).
The dot product is 3+4+3=10, and both vectors have magnitude sqrt14, so cos(theta)=10/14=5/7, giving theta=cos^-1(5/7).
✦ theta=cos^-1(5/7)
Find the projection of the vector i-j on the vector i+j.
Hint. Use the projection formula: projection of a on b is (a.b)/|b|.
Since (i-j).(i+j)=1-1=0, the projection is 0/sqrt2=0, meaning the two vectors are perpendicular.
✦ 0 (the vectors are perpendicular)
Find the projection of the vector i+3j+7k on the vector 7i-j+8k.
Hint. Use the projection formula: projection of a on b is (a.b)/|b|.
Since (i+3j+7k).(7i-j+8k)=7-3+56=60, and |7i-j+8k|=sqrt114, the projection is 60/sqrt114.
✦ 60/sqrt(114)
Show that each of the given three vectors (1/7)(2i+3j+6k), (1/7)(3i-6j+2k), (1/7)(6i+2j-3k) is a unit vector. Also, show that they are mutually perpendicular to each other.
Hint. Compute each vector's magnitude to confirm it is 1, then compute each pairwise dot product to confirm it is 0.
Since each vector's squared magnitude is (4+9+36)/49=1 (and similarly for the other two by symmetry of the sum of squares), all three are unit vectors; because each pairwise dot product (e.g. (2)(3)+(3)(-6)+(6)(2)=6-18+12=0, scaled by 1/49) works out to zero, this also confirms mutual perpendicularity.
✦ All three are unit vectors and mutually perpendicular, as verified by direct computation
Find |a| and |b|, if (a+b).(a-b)=8 and |a|=8|b|.
Hint. Expand the dot product using the distributive law: (a+b).(a-b)=|a|^2-|b|^2.
Since (a+b).(a-b)=|a|^2-|b|^2=8, and |a|=8|b| gives |a|^2=64|b|^2, substituting yields 64|b|^2-|b|^2=8, so 63|b|^2=8, giving |b|^2=8/63.
✦ |b|=sqrt(8/63), |a|=8.sqrt(8/63)
Evaluate the product (3a-5b).(2a+7b).
Hint. Expand using the distributive law, keeping a.a=|a|^2, b.b=|b|^2, and a.b=b.a.
Expanding term by term gives 6(a.a)+21(a.b)-10(b.a)-35(b.b)=6|a|^2+11(a.b)-35|b|^2, since the two cross terms combine as 21-10=11 times a.b.
✦ 6|a|^2+11(a.b)-35|b|^2
Find the magnitude of two vectors a and b, having the same magnitude and such that the angle between them is 60 degrees and their scalar product is 1/2.
Hint. Use a.b=|a||b|cos(theta) with |a|=|b| to set up a single equation for the common magnitude.
Since |a|=|b|=r and the angle is 60 degrees, a.b=r^2.cos(60)=r^2/2=1/2, so r^2=1, giving r=1.
✦ |a|=|b|=1
Find |x|, if for a unit vector a, (x-a).(x+a)=12.
Hint. Expand the dot product: (x-a).(x+a)=|x|^2-|a|^2.
Since a is a unit vector, |a|^2=1, so |x|^2-1=12, giving |x|^2=13.
✦ |x|=sqrt(13)
If a=2i+2j+3k, b=-i+2j+k and c=3i+j are such that a+lambda.b is perpendicular to c, then find the value of lambda.
Hint. Set the dot product of (a+lambda.b) with c equal to zero and solve for lambda.
Expanding (a+lambda.b).c=a.c+lambda(b.c)=0: since a.c=6+2+0=8 and b.c=-3+2+0=-1, solving 8-lambda=0 gives lambda=8.
✦ lambda=8
Show that |a|b+|b|a is perpendicular to |a|b-|b|a, for any two nonzero vectors a and b.
Hint. Compute the dot product of the two given combinations and simplify using a.b=b.a.
Expanding the dot product (|a|b+|b|a).(|a|b-|b|a)=|a|^2(b.b)-|b|^2(a.a)=|a|^2|b|^2-|b|^2|a|^2=0, confirming the two vectors are perpendicular.
✦ Proved: the dot product simplifies to exactly 0
If a.a=0 and a.b=0, then what can be concluded about the vector b?
Hint. Recall that a.a=0 forces a itself to be the zero vector.
Since a.a=|a|^2=0 forces a to be the zero vector, the condition a.b=0 is automatically satisfied for any vector b, so no conclusion can be drawn about b at all.
✦ No conclusion can be drawn about b (since a.a=0 already forces a=0, making a.b=0 automatic)
If a, b, c are unit vectors such that a+b+c=0, find the value of a.b+b.c+c.a.
Hint. Square both sides of a+b+c=0 (i.e. dot the equation with itself) and expand.
Squaring a+b+c=0 gives (a+b+c).(a+b+c)=0, which expands to |a|^2+|b|^2+|c|^2+2(a.b+b.c+c.a)=0; since each vector is a unit vector, 3+2(a.b+b.c+c.a)=0, giving the value.
✦ a.b+b.c+c.a=-3/2
If either vector a=0 or b=0, then a.b=0. But the converse need not be true. Justify your answer with an example.
Hint. Find two nonzero, mutually perpendicular vectors whose dot product is zero despite neither being the zero vector.
Taking a=i and b=j, both are nonzero vectors, yet a.b=(1)(0)+(0)(1)+(0)(0)=0, since the vectors are perpendicular rather than either being zero, showing the converse fails.
✦ Example: a=i, b=j; both nonzero but a.b=0, since they are perpendicular
If the vertices A, B, C of a triangle ABC are (1,2,3), (-1,0,0), (0,1,2) respectively, then find angle ABC. [Angle ABC is the angle between the vectors BA and BC].
Hint. Form the vectors BA and BC from the given coordinates, then apply the angle formula.
With BA=(2,2,3) and BC=(1,1,2), the dot product is 2+2+6=10, and the magnitudes are sqrt17 and sqrt6, so cos(angle ABC)=10/sqrt102.
✦ angle ABC=cos^-1(10/sqrt(102))
Show that the points A(1,2,7), B(2,6,3) and C(3,10,-1) are collinear.
Hint. Form vectors AB and AC and check whether one is a scalar multiple of the other.
Since AB=(1,4,-4) and AC=(2,8,-8)=2.AB, the vector AC is exactly twice AB, so the three points lie on the same straight line.
✦ Collinear, since AC=2.AB
Show that the vectors 2i-j+k, i-3j-5k and 3i-4j-4k form the vertices of a right angled triangle.
Hint. Compute the squared lengths of all three sides (differences between the given position vectors) and check the Pythagorean relation.
Labelling the vectors as a, b, c and computing |a-b|^2=41, |b-c|^2=6, |c-a|^2=35, the relation 6+35=41 holds, confirming a right angle at the vertex between the two shorter sides.
✦ Right angled triangle, since 6+35=41
Choose the correct answer: if a is a nonzero vector of magnitude a and lambda a nonzero scalar, then lambda.a is a unit vector if (A) lambda=1 (B) lambda=-1 (C) a=|lambda| (D) a=1/|lambda|
Hint. The magnitude of lambda.a is |lambda|.a; set this equal to 1 and solve for the relationship between a and lambda.
Since |lambda.a|=|lambda|.a, requiring this to equal 1 gives a=1/|lambda|, matching exactly one option regardless of the sign of lambda.
✦ (D) a=1/|lambda|
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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