Find |a x b|, if a=i-7j+7k and b=3i-2j+2k.
Hint. Compute the cross product using the determinant formula, then find its magnitude.
Expanding the determinant with rows i,j,k / 1,-7,7 / 3,-2,2 gives a x b=(-14+14)i-(2-21)j+(-2+21)k=0i+19j+19k, so the magnitude is sqrt(0+361+361).
✦ |a x b|=19.sqrt(2)
Find a unit vector perpendicular to each of the vectors a+b and a-b, where a=3i+2j+2k and b=i+2j-2k.
Hint. Compute a+b and a-b first, then take their cross product and normalise it.
With a+b=4i+4j+0k and a-b=2i+0j+4k, expanding the determinant gives their cross product as 16i-16j-8k, which has magnitude sqrt(256+256+64)=24, so dividing through by 24 gives the unit vector.
✦ (2/3)i-(2/3)j-(1/3)k
If a unit vector a makes angles pi/3 with i, pi/4 with j and an acute angle theta with k, then find theta and hence, the components of a.
Hint. Use the fact that the sum of the squares of the direction cosines of a unit vector equals 1.
Since cos^2(pi/3)+cos^2(pi/4)+cos^2(theta)=1, substituting 1/4+1/2+cos^2(theta)=1 gives cos^2(theta)=1/4, so cos(theta)=1/2 (taking the acute angle), giving theta=pi/3.
✦ theta=pi/3; components are (1/2)i+(1/sqrt2)j+(1/2)k
Show that (a-b) x (a+b)=2(a x b).
Hint. Expand the cross product using distributivity, and use a x a=0=b x b.
Expanding, (a-b)x(a+b)=a x a+a x b-b x a-b x b=0+a x b+a x b-0, using b x a=-(a x b), which simplifies to 2(a x b).
✦ Proved: (a-b) x (a+b)=2(a x b)
Find lambda and mu if (2i+6j+27k) x (i+lambda.j+mu.k)=0.
Hint. The cross product is zero exactly when the two vectors are parallel, so their components must be proportional.
Since the cross product vanishes only for parallel vectors, the ratios of corresponding components must match: 2/1=6/lambda=27/mu, giving lambda and mu directly.
✦ lambda=3, mu=27/2
Given that a.b=0 and a x b=0. What can you conclude about the vectors a and b?
Hint. Recall what each condition individually implies, then combine them.
Since a.b=0 means a and b are perpendicular (or one is zero) while a x b=0 means they are parallel (or one is zero), and a nonzero vector cannot be simultaneously perpendicular and parallel to another nonzero vector, at least one of a or b must be the zero vector.
✦ Either a=0 or b=0 (or both)
Let the vectors a, b, c be given as a1.i+a2.j+a3.k, b1.i+b2.j+b3.k, c1.i+c2.j+c3.k. Then show that a x (b+c)=a x b+a x c.
Hint. Write out the cross product in determinant form for both sides and compare component by component.
Expanding a x (b+c) using the determinant formula with (b1+c1, b2+c2, b3+c3) as the second row, and separately expanding a x b and a x c, shows each component matches exactly since the cross product's determinant expansion is linear in each row.
✦ Proved: cross product distributes over vector addition
If either a=0 or b=0, then a x b=0. Is the converse true? Justify your answer with an example.
Hint. Find two nonzero, parallel (collinear) vectors whose cross product is zero despite neither being the zero vector.
Taking a=i and b=2i, both are nonzero vectors, yet a x b=0 since they are parallel rather than either being zero, showing the converse fails.
✦ Not true. Example: a=i, b=2i; both nonzero but a x b=0, since they are parallel
Find the area of the triangle with vertices A(1,1,2), B(2,3,5) and C(1,5,5).
Hint. Form two side vectors from one vertex, take their cross product, and halve its magnitude.
With AB=(1,2,3) and AC=(0,4,3), the cross product is (-6,-3,4), with magnitude sqrt61, so the triangle's area is half of that.
✦ Area=sqrt(61)/2 square units
Find the area of the parallelogram whose adjacent sides are determined by the vectors a=i-j+3k and b=2i-7j+k.
Hint. The area of the parallelogram is the magnitude of the cross product of its two adjacent side vectors.
The cross product of a and b is (20,5,-5), with magnitude sqrt450=15sqrt2, which is exactly the parallelogram's area.
✦ Area=15.sqrt(2) square units
Choose the correct answer: let the vectors a and b be such that |a|=3 and |b|=sqrt2/3, then a x b is a unit vector, if the angle between a and b is (A) pi/6 (B) pi/4 (C) pi/3 (D) pi/2
Hint. Set |a x b|=|a||b|sin(theta) equal to 1 and solve for theta.
Since |a x b|=3.(sqrt2/3).sin(theta)=sqrt2.sin(theta), setting this equal to 1 gives sin(theta)=1/sqrt2, so theta=pi/4.
✦ (B) pi/4
Choose the correct answer: area of a rectangle having vertices A, B, C and D with position vectors -i+(1/2)j+4k, i+(1/2)j+4k, i-(1/2)j+4k and -i-(1/2)j+4k, respectively is (A) 1/2 (B) 1 (C) 2 (D) 4
Hint. Find two adjacent side vectors of the rectangle and take the magnitude of their cross product.
Since AB=2i (from A to B) and AD=-j (from A to D), the cross product AB x AD has magnitude 2.1=2, since the two sides are perpendicular and their lengths are 2 and 1 respectively.
✦ (C) 2
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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