Compute the magnitude of the following vectors: a=i+j+k; b=2i-7j-3k; c=(1/sqrt3)i+(1/sqrt3)j-(1/sqrt3)k.
Hint. Use |v|=sqrt(x^2+y^2+z^2) for each vector.
Since magnitude is the square root of the sum of squared components, a gives sqrt(1+1+1)=sqrt3, b gives sqrt(4+49+9)=sqrt62, and c gives sqrt(1/3+1/3+1/3)=1, which means c is already a unit vector.
✦ |a|=sqrt(3), |b|=sqrt(62), |c|=1
Write two different vectors having the same magnitude.
Hint. Choose any two vectors whose components give the same value of sqrt(x^2+y^2+z^2) but point in different directions.
Since magnitude only depends on sqrt(x^2+y^2+z^2), any two vectors with components that are permutations or sign-changes of each other (but not identical) will share a magnitude while pointing differently.
✦ i+2j+2k and 2i+2j+k both have magnitude 3
Write two different vectors having the same direction.
Hint. Choose one vector, then scale it by any two different positive constants — both results point the same way.
Since scaling a vector by a positive constant preserves its direction, any two positive multiples of the same base vector are different vectors pointing the same way.
✦ i+2j+3k and 2i+4j+6k both point in the same direction
Find the values of x and y so that the vectors 2i+3j and xi+yj are equal.
Hint. Two vectors are equal exactly when their corresponding components match.
Since equal vectors have identical components, matching the i and j coefficients directly gives the values of x and y.
✦ x=2, y=3
Find the scalar and vector components of the vector with initial point (2,1) and terminal point (-5,7).
Hint. Subtract the initial point's coordinates from the terminal point's coordinates to get the vector's components.
Since the vector equals (terminal point) minus (initial point), subtracting gives the scalar components -7 and 6, and the vector component is the corresponding combination of i and j.
✦ Scalar components: -7, 6. Vector component: -7i+6j
Find the sum of the vectors a=i-2j+k, b=-2i+4j+5k and c=i-6j-7k.
Hint. Add the corresponding i, j, and k components of all three vectors separately.
Adding the i-components (1-2+1=0), j-components (-2+4-6=-4), and k-components (1+5-7=-1) separately gives the resultant vector.
✦ a+b+c=-4j-k
Find the unit vector in the direction of the vector a=i+j+2k.
Hint. Divide the vector by its own magnitude.
Since |a|=sqrt(1+1+4)=sqrt6, dividing each component by sqrt6 gives the unit vector in the same direction.
✦ (1/sqrt6)i+(1/sqrt6)j+(2/sqrt6)k
Find the unit vector in the direction of vector PQ, where P and Q are the points (1,2,3) and (4,5,6) respectively.
Hint. First find the vector PQ by subtracting P from Q, then divide by its magnitude.
Since PQ=Q-P=3i+3j+3k, with magnitude sqrt(27)=3sqrt3, dividing each component by 3sqrt3 gives the unit vector.
✦ (1/sqrt3)i+(1/sqrt3)j+(1/sqrt3)k
For given vectors, a=2i-j+2k and b=-i+j-k, find the unit vector in the direction of the vector a+b.
Hint. First add the two vectors, then divide the sum by its own magnitude.
Since a+b=i+0j+k, with magnitude sqrt2, dividing each component by sqrt2 gives the unit vector in the direction of the sum.
✦ (1/sqrt2)i+(1/sqrt2)k
Find a vector in the direction of vector 5i-j+2k which has magnitude 8 units.
Hint. Find the unit vector in that direction first, then scale it up to magnitude 8.
Since the given vector has magnitude sqrt(25+1+4)=sqrt30, scaling it by 8/sqrt30 gives a vector in the same direction with magnitude exactly 8.
✦ (40/sqrt30)i-(8/sqrt30)j+(16/sqrt30)k
Show that the vectors 2i-3j+4k and -4i+6j-8k are collinear.
Hint. Check whether one vector is a scalar multiple of the other.
Since -4i+6j-8k=-2(2i-3j+4k), the second vector is exactly -2 times the first, so by definition they are collinear (parallel).
✦ Collinear, since the second vector equals -2 times the first
Find the direction cosines of the vector i+2j+3k.
Hint. Divide each component by the vector's magnitude to get the direction cosines.
Since the magnitude is sqrt(1+4+9)=sqrt14, dividing each component by sqrt14 gives the three direction cosines.
✦ (1/sqrt14, 2/sqrt14, 3/sqrt14)
Find the direction cosines of the vector joining the points A(1,2,-3) and B(-1,-2,1), directed from A to B.
Hint. First find vector AB by subtracting A from B, then find its direction cosines.
Since AB=B-A=-2i-4j+4k, with magnitude sqrt(4+16+16)=6, dividing each component by 6 gives the direction cosines.
✦ (-1/3, -2/3, 2/3)
Show that the vector i+j+k is equally inclined to the axes OX, OY and OZ.
Hint. Compute the direction cosines and check whether all three are equal.
Since the magnitude of i+j+k is sqrt3, its direction cosines are all 1/sqrt3, and because all three direction cosines are equal, the vector makes the same angle with all three axes.
✦ Direction cosines are all 1/sqrt3, so the angles with OX, OY, OZ are all equal
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are i+2j-k and -i+j+k respectively, in the ratio 2:1, (i) internally (ii) externally.
Hint. Apply the section formula: internal division uses (m.Q+n.P)/(m+n), external uses (m.Q-n.P)/(m-n).
With P=i+2j-k, Q=-i+j+k, m=2, n=1: internally, R=(2Q+P)/3=(-i+4j+k)/3; externally, R=(2Q-P)/1=-3i+3k.
✦ (i) (-1/3)i+(4/3)j+(1/3)k (ii) -3i+3k
Find the position vector of the mid point of the vector joining the points P(2,3,4) and Q(4,1,-2).
Hint. The midpoint's position vector is the average of the two position vectors.
Since the midpoint is (P+Q)/2, averaging the corresponding coordinates (2+4)/2, (3+1)/2, (4-2)/2 gives the midpoint's position vector.
✦ 3i+2j+k
Show that the points A, B and C with position vectors a=3i-4j-4k, b=2i-j+k and c=i-3j-5k, respectively form the vertices of a right angled triangle.
Hint. Compute the squared lengths of all three sides and check whether the Pythagorean relation holds.
Computing AB=b-a, BC=c-b, CA=a-c and their squared magnitudes gives 35, 41, and 6 respectively; since 35+6=41, the Pythagorean relation holds for the two shorter sides, confirming a right angle at the vertex between them.
✦ Right angled triangle, since |AB|^2+|CA|^2=|BC|^2 (35+6=41)
Choose the correct answer: in triangle ABC (Fig 10.18), which of the following is not true? (A) AB+BC+CA=0 (B) AB+BC-AC=0 (C) AB+BC-CA=0 (D) AB-CB+CA=0
Hint. Use the triangle law AB+BC=AC to check each option, remembering that CA=-AC and CB=-BC.
Since AB+BC=AC always holds, options (A), (B), and (D) all reduce to true identities once CA and CB are rewritten as -AC and -BC; only option (C), which effectively requires AC=-AC, fails to hold in general.
✦ (C) AB+BC-CA=0 is not true
Choose the correct answer: if a and b are two collinear vectors, then which of the following are incorrect? (A) b=lambda.a for some scalar lambda (B) a=+-b (C) the respective components of a and b are not proportional (D) both vectors a and b have same direction, but different magnitudes.
Hint. Recall that collinear vectors always have proportional components by definition.
Since collinearity means one vector is a scalar multiple of the other, their respective components must always be proportional, making statement (C) incorrect (a false claim about collinear vectors), while (A) is the very definition and (D) is a valid special case.
✦ (C) is incorrect
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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