JEE Mathematics Vector Algebra MCQs Set 31

Answer: (a) \( \frac{\vec b+2\vec c}{3}-\vec a \)

Explanation:
1. Section formula for BD : DC = 2 : 1: \( \vec d=\frac{1\cdot\vec b+2\cdot\vec c}{3} \).
2. Hence \( \overrightarrow{AD}=\vec d-\vec a=\frac{\vec b+2\vec c}{3}-\vec a \).

Teacher's Note:
1. In ratio \( m \) : \( n \) from \( B \) to \( C \) the position vector is (\( n \) times \( B + m \) times \( C)/(m+n \)).
2. Always subtract the position vector of the starting point to get a displacement.

Answer: (c) \( 5\hat i+\hat j+4\hat k \) and \( 4\hat i-\hat j+5\hat k \)

Explanation:
1. \( D \) divides BC as 1 : 2 from \( B \): \( \overrightarrow{AD}=\frac{2\overrightarrow{AB}+\overrightarrow{AC}}{3}=\frac{(12,6,6)+(3,-3,6)}{3}=(5,1,4) \).
2. \( D' \) divides BC as 2 : 1: \( \overrightarrow{AD'}=\frac{\overrightarrow{AB}+2\overrightarrow{AC}}{3}=\frac{(6,3,3)+(6,-6,12)}{3}=(4,-1,5) \).

Teacher's Note:
1. Use \( A \) as origin so that AB and AC act as position vectors of \( B \) and \( C \).
2. Nearer to \( B \) means \( B \) gets the larger weight.

Answer: (a) \( 2\sqrt6 \)

Explanation:
1. Add: \( \vec b+\vec c=(2,4,2) \).
2. Projection on \( \vec a \) is \( \frac{(2,4,2)\cdot(1,2,1)}{\sqrt6}=\frac{12}{\sqrt6}=2\sqrt6 \).

Teacher's Note:
1. Projection of \( u \) on a = (u dot a)/|a|.
2. Add the vectors first, then project once.

Answer: (a) \( \frac{7}{\sqrt3} \)

Explanation:
1. Segment vector \( (2,3,2) \). Equal acute angles with the axes give direction cosines \( \left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right) \).
2. Projection \( =\frac{2+3+2}{\sqrt3}=\frac7{\sqrt3} \).

Teacher's Note:
1. Equal angles with the axes means \( l = m = n = 1/sqrt 3 \).
2. Projection of a vector on a line = dot product with the unit direction.

Answer: (d) \( 120^\circ \)

Explanation:
1. Fourth vertex \( D=A+C-B=(-1,1,1) \).
2. So \( \overrightarrow{AC}=(8,2,2) \) and \( \overrightarrow{BD}=(-4,4,0) \).
3. Dot product \( -32+8=-24 \); magnitudes \( 6\sqrt2 \) and \( 4\sqrt2 \).
4. Hence \( \cos\theta=\frac{-24}{48}=-\frac12 \) and \( \theta=120^\circ \).

Teacher's Note:
1. In a parallelogram ABCD, \( D = A + C - B \).
2. The angle between two vectors is taken from the dot product, so an obtuse value is allowed.

Answer: (c) \( -3\hat i+3\hat j+3\hat k \)

Explanation:
1. The cross product gives the common perpendicular: \( (2,3,-1)\times(1,-2,3)=(7,-7,-7) \), so \( \vec c=t(1,-1,-1) \).
2. Condition: \( t(2+1-1)+6=0 \), so \( t=-3 \).
3. Hence \( \vec c=(-3,3,3) \).

Teacher's Note:
1. A vector perpendicular to two vectors is parallel to their cross product.
2. Fix the scalar multiple using the extra condition.

Answer: (c) \( 5 \)

Explanation:
1. Triangles ABC and BCD each have area 6. \( O \) is the centroid of triangle ABC (AM and BO are both medians), so area of triangle BOM = \( \frac16\times6=1 \).
2. Quadrilateral OMCD = triangle BCD - triangle BOM = \( 6-1=5 \).

Teacher's Note:
1. A median splits a triangle into two equal areas and the three medians make six equal parts.
2. Subtract the small triangle from the triangle BCD.

Answer: (c) \( \frac12 \)

Explanation:
1. Edges from \( P \): \( \overrightarrow{PQ}=(-1,-2,-1),\ \overrightarrow{PR}=(1,-5,-1),\ \overrightarrow{PS}=(5,-2,2) \).
2. Determinant \( =-1(-10-2)+2(2+5)-1(-2+25)=12+14-23=3 \).
3. Volume \( =\frac16|3|=\frac12 \).

Teacher's Note:
1. Volume of a tetrahedron = one sixth of the absolute scalar triple product of three edges from a vertex.
2. Expand the determinant along the first row.

Answer: (c) \( 1 \)

Explanation:
1. Linear dependence of three vectors in space means the determinant is 0.
2. Expand: \( 1(3x-4)+2(-2x+4)+3(2-3)=3x-4-4x+8-3=1-x \).
3. So \( x=1 \).

Teacher's Note:
1. Three vectors are dependent iff they are coplanar iff their determinant vanishes.
2. Expand along the first row.

Answer: (a) \( (0,10,-3) \)

Explanation:
1. Use \( \vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c \).
2. Here \( \vec a\cdot\vec c=3+4+3=10 \) and \( \vec a\cdot\vec b=2-2+3=3 \).
3. So \( p=0,\ q=10,\ r=-3 \).

Teacher's Note:
1. BAC-CAB rule gives the result directly as a combination of \( b \) and \( c \).
2. Compute the two dot products only.

Answer: (c) The four points are coplanar

Explanation:
1. Edges from the first point: \( (-2,-4,-6),\ (-1,-3,-8),\ (1,0,-7) \).
2. Determinant \( -2(21-0)+4(7+8)-6(0+3)=-42+60-18=0 \).
3. Zero scalar triple product means the points are coplanar.

Teacher's Note:
1. Four points are coplanar iff the scalar triple product of three edges from one point is zero.
2. A zero determinant also means zero volume.

Answer: (a) parallel

Explanation:
1. Compute \( (\vec a-\vec d)\times(\vec b-\vec c)=\vec a\times\vec b-\vec a\times\vec c-\vec d\times\vec b+\vec d\times\vec c \).
2. Using the given relations: \( \vec c\times\vec d-\vec b\times\vec d+\vec b\times\vec d-\vec c\times\vec d=\vec 0 \).
3. Zero cross product means the two vectors are parallel.

Teacher's Note:
1. A zero cross product of two non-zero vectors means they are parallel.
2. Expand and substitute both given equalities.

Answer: (b) \( \vec 0 \)

Explanation:
1. Non-coplanar vectors span all of space, so any vector can be written as \( \vec v=x\vec a+y\vec b+z\vec c \).
2. Then \( |\vec v|^2=\vec v\cdot(x\vec a+y\vec b+z\vec c)=0 \), so \( \vec v=\vec 0 \).

Teacher's Note:
1. Three non-coplanar vectors form a basis of space.
2. A vector perpendicular to every basis vector has zero length.

Answer: (a) \( x\in(-3,2) \)

Explanation:
1. Cyclic property: \( [\vec b\,\vec c\,\vec a]=[\vec a\,\vec b\,\vec c]=\begin{vmatrix}x&12&-1\\2&2x&1\\1&0&1\end{vmatrix} \).
2. Expand: \( x(2x)-12(2-1)-1(0-2x)=2x^2+2x-12=2(x+3)(x-2) \).
3. Left-handed means the value is negative: \( -3<x<2 \).

Teacher's Note:
1. A right-handed triple has positive scalar triple product, a left-handed one negative.
2. Rotating the order cyclically does not change the value.

Answer: (d) \( \sqrt{57} \)

Explanation:
1. Parallelogram law: \( |\vec a-\vec b|^2+|\vec a+\vec b|^2=2(|\vec a|^2+|\vec b|^2) \).
2. So \( 64+100=2(25+|\vec b|^2) \), giving \( |\vec b|^2=57 \) and \( |\vec b|=\sqrt{57} \).

Teacher's Note:
1. Add the squares of sum and difference to cancel the dot product.
2. Remember to take the square root at the end.

Answer: (d) \( \sin^{-1}\sqrt{\frac89} \)

Explanation:
1. Diagonals: \( \vec a+\vec b=(3,0,0) \) and \( \vec a-\vec b=(1,2,2) \).
2. Dot product 3; magnitudes 3 and 3; \( \cos\theta=\frac13 \).
3. Then \( \sin\theta=\sqrt{1-\frac19}=\sqrt{\frac89} \), so \( \theta=\sin^{-1}\sqrt{8/9} \).

Teacher's Note:
1. The diagonals are \( a + b \) and \( a - b \).
2. Convert \( \cos = 1/3 \) into the sine form given in the options.

Answer: (a) \( 16 \)

Explanation:
1. Use \( |\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2 \).
2. So \( |\vec a\times\vec b|^2=400-144=256 \) and \( |\vec a\times\vec b|=16 \).

Teacher's Note:
1. Lagrange's identity links dot and cross products.
2. Take the positive square root.

Answer: (d) 21 : 4

Explanation:
1. Take \( P \) as origin, \( Q = \vec q \), \( S = \vec s \), \( R = \vec q+\vec s \).
2. \( X = \vec q+\tfrac45\vec s \) and \( Y = \tfrac15\vec q+\vec s \).
3. \( Z = t(\vec q+\vec s)=X+u(Y-X) \): comparing coefficients, \( t=1-\tfrac45u \) and \( t=\tfrac45+\tfrac15u \), so \( u=\tfrac15,\ t=\tfrac{21}{25} \).
4. Hence PZ : ZR = 21 : 4.

Teacher's Note:
1. Choose \( P \) as origin and the two sides as base vectors.
2. Equate the coefficients of the independent vectors \( q \) and \( s \).

Answer: (d) \( \vec a-\vec b,\ \vec b-\vec c,\ \vec c-\vec a \)

Explanation:
1. The three differences add up to zero: \( (\vec a-\vec b)+(\vec b-\vec c)+(\vec c-\vec a)=\vec 0 \), so this set is dependent.
2. The other sets have non-zero determinants: \( 2 \) for the first sum set, \( -6 \) for the third and \( [\vec a\,\vec b\,\vec c]^2\neq0 \) for the cross products.

Teacher's Note:
1. A set is dependent when some combination with non-zero coefficients vanishes.
2. Check coefficient determinants for the other sets.

Answer: (c) \( \frac{c\vec A-(\vec A\times\vec B)}{|\vec A|^2} \)

Explanation:
1. Cross with \( \vec A \): \( \vec A\times(\vec A\times\vec x)=(\vec A\cdot\vec x)\vec A-|\vec A|^2\vec x=\vec A\times\vec B \).
2. Substitute \( \vec A\cdot\vec x=c \): \( \vec x=\frac{c\vec A-\vec A\times\vec B}{|\vec A|^2} \).

Teacher's Note:
1. Take the cross product of the second equation with \( A \).
2. Use the triple product expansion and the first equation.

Answer: (a) \( \frac\pi2 \) and \( \frac\pi3 \)

Explanation:
1. Expand: \( 2[(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c]=\vec b \).
2. Since \( b \) and \( c \) are independent: \( 2\,\vec a\cdot\vec c=1 \) and \( \vec a\cdot\vec b=0 \).
3. So the angle with \( b \) is \( \frac\pi2 \) and with \( c \) is \( \frac\pi3 \).

Teacher's Note:
1. Compare coefficients of the independent vectors \( b \) and \( c \).
2. For unit vectors the dot product equals the cosine.

Answer: (c) \( 3 \)

Explanation:
1. Here \( \overrightarrow{AB}=(20,-4,-6) \) and \( \overrightarrow{AN}=(11,0,-1) \).
2. Cross product \( \overrightarrow{AN}\times\overrightarrow{AB}=(-4,46,-44) \), with squared magnitude \( 16+2116+1936=4068 \).
3. Also \( |\overrightarrow{AB}|^2=452 \). Distance \( =\sqrt{\frac{4068}{452}}=\sqrt9=3 \).

Teacher's Note:
1. Distance from a point to a line = |(point - A) x direction| / |direction|.
2. Use any point \( A \) on the line.

Answer: (c) \( \frac{a+b+c}{3} \)

Explanation:
1. The join of \( (a,b,c) \) and \( (r,r,r) \) is perpendicular to \( (1,1,1) \): \( (a-r)+(b-r)+(c-r)=0 \).
2. So \( r=\frac{a+b+c}{3} \).

Teacher's Note:
1. The line \( x = y = z \) has direction (1, 1, 1).
2. Set the dot product of the join with the direction equal to zero.

Answer: (d) a pair of perpendicular planes

Explanation:
1. Factorise: \( y(x+z)=0 \), so \( y=0 \) or \( x+z=0 \).
2. The normals \( (0,1,0) \) and \( (1,0,1) \) have dot product 0, so the planes are perpendicular.

Teacher's Note:
1. In three dimensions an equation is a surface, so the factors are planes.
2. Check the dot product of the two normals.

Answer: (a) The line is parallel to \( 2\hat i+6\hat j \)

Explanation:
1. Direction is \( \hat i-3\hat j \), not \( 2\hat i+6\hat j \) (they are not proportional), so the first statement is wrong.
2. At \( \lambda=1 \) the point is \( (3,3) \) and at \( \lambda=-1 \) it is \( (1,9) \), so the next two are correct.
3. The z-component never changes, so the line is parallel to the xy-plane.

Teacher's Note:
1. Compare direction ratios for parallelism, not position vectors.
2. Test points by choosing suitable values of \( \lambda \).

Answer: (d) \( 5\sqrt2 \)

Explanation:
1. Adding the three conditions: \( \vec a\cdot\vec b+\vec a\cdot\vec c=0 \), etc., so \( \vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a=0 \).
2. Then \( |\vec a+\vec b+\vec c|^2=9+16+25+0=50 \).
3. Hence the magnitude is \( 5\sqrt2 \).

Teacher's Note:
1. Add the three perpendicularity conditions to get the sum of the pairwise dot products.
2. Square the sum of the vectors.

Answer: (b) \( 6 \)

Explanation:
1. From \( \vec a\times(\vec b-2\vec c)=\vec 0 \): \( \vec b-2\vec c=t\vec a \), so \( \ell=2,\ \mu=t \).
2. Dot with \( \vec c \): \( \vec b\cdot\vec c=4\cdot\frac14=1=2+t\,\vec a\cdot\vec c \), so \( t\,\vec a\cdot\vec c=-1 \).
3. Magnitude: \( 16=4+t^2+4t\,\vec a\cdot\vec c=t^2 \), so \( |t|=4 \).
4. Thus \( |\ell|+|\mu|=2+4=6 \).

Teacher's Note:
1. A zero cross product means the vectors are parallel.
2. Use two dot-product conditions to find the scalar.

Answer: (a) \( \left[0,\frac\pi6\right) \)

Explanation:
1. For unit vectors \( |\vec e_1-\vec e_2|=2\left|\sin\theta\right| \).
2. Condition \( 2\sin\theta<1 \) gives \( \sin\theta<\frac12 \).
3. In \( [0,\pi] \) this holds for \( \theta\in\left[0,\frac\pi6\right)\cup\left(\frac{5\pi}6,\pi\right] \), so the interval \( \left[0,\frac\pi6\right) \) is a possibility.

Teacher's Note:
1. \( |e1 - e2|^2 = 2 - 2 \cos(2 \theta) = 4 \sin^2 \theta \).
2. Solve the inequality over the stated range of \( \theta \).

Answer: (c) \( \frac43 \)

Explanation:
1. Take the cube with vertices at \( (0,0,0) \) to \( (1,1,1) \); diagonals have directions \( \frac{(\pm1,\pm1,1)}{\sqrt3} \).
2. For a unit vector \( (l,m,n) \) the four cosines are \( \frac{l+m+n}{\sqrt3},\frac{-l+m+n}{\sqrt3},\frac{l-m+n}{\sqrt3},\frac{l+m-n}{\sqrt3} \).
3. Sum of squares \( =\frac{4(l^2+m^2+n^2)}{3}=\frac43 \).

Teacher's Note:
1. Write each diagonal direction and take the dot product with a general unit vector.
2. Cross terms cancel when the four squares are added.

Answer: (a) coplanar but cannot form a triangle

Explanation:
1. The third vector equals the sum of the first two: \( (1,2,3)+(2,-1,1)=(3,1,4) \), so the three are coplanar.
2. To form a triangle placed head-to-tail the sum must be zero, but here the sum is \( 2(3,1,4)\ne\vec 0 \).
3. So they are coplanar but cannot form a triangle.

Teacher's Note:
1. Three vectors form a closed triangle only if their sum is zero.
2. Coplanarity only needs one to be a combination of the others.