JEE Mathematics Vector Algebra MCQs Set 24

Answer: (a) \( \vec a_1+\vec a_2+\vec a_3+\vec a_4=\vec 0 \)

Explanation:
1. Take the vertex at the origin with edges \( \vec p,\vec q,\vec r \). The outward area vectors of the three faces through the origin are \( \tfrac12\vec q\times\vec p,\ \tfrac12\vec r\times\vec q,\ \tfrac12\vec p\times\vec r \) and the fourth is \( \tfrac12(\vec q-\vec p)\times(\vec r-\vec p) \).
2. Adding all four gives the zero vector (the vector area of a closed surface is zero).

Teacher's Note:
1. The vector area of any closed surface is zero.
2. Check by writing the four area vectors with cross products.

Answer: (b) \( 0 \)

Explanation:
1. Dot with \( \vec a+\vec b+\vec c \): \( \vec r\cdot(\vec a+\vec b+\vec c)=\ell[\vec a\,\vec b\,\vec c]+m[\vec a\,\vec b\,\vec c]+n[\vec a\,\vec b\,\vec c] \).
2. This is zero, and the triple product is 2, so \( \ell+m+n=0 \).

Teacher's Note:
1. Each cross product is perpendicular to two of the vectors, so most terms vanish when dotted.
2. The surviving terms all carry the same triple product.

Answer: (a) \( a^2(\vec b\cdot\vec c) \)

Explanation:
1. Use \( (\vec a\times\vec b)\cdot(\vec a\times\vec c)=a^2(\vec b\cdot\vec c)-(\vec a\cdot\vec c)(\vec a\cdot\vec b) \).
2. Since \( \vec a \) is perpendicular to both \( \vec b \) and \( \vec c \), \( \vec a\cdot\vec b=\vec a\cdot\vec c=0 \).
3. Hence the value is \( a^2(\vec b\cdot\vec c) \).

Teacher's Note:
1. Apply the Lagrange identity for the dot product of two cross products.
2. A vector parallel to \( b x c \) is perpendicular to \( b \) and \( c \).

Answer: (d) (2) and (4) only

Explanation:
1. Orthogonality: \( \tan^2\alpha-\tan\alpha-6=0 \), so \( \tan\alpha=3 \) or \( -2 \).
2. Obtuse with the z-axis: \( \sin2\alpha<0 \). For \( \tan\alpha=3 \), \( \sin2\alpha=\frac{2\cdot3}{1+9}>0 \) - rejected.
3. For \( \tan\alpha=-2 \), \( \sin2\alpha=-\frac45<0 \) - accepted, giving \( \alpha=\pi-\tan^{-1}2 \) or \( 2\pi-\tan^{-1}2 \).

Teacher's Note:
1. Use the dot product of \( b \) and \( c \), noting \( \sqrt{\sin} \) times \( \sqrt{\csc} \) is 1.
2. The sign of the z-component decides whether the angle with the z-axis is obtuse.

Answer: (b) \( 2 \)

Explanation:
1. Magnitude doubled: \( 16+(4x-2)^2+4=4(10+x^2) \) gives \( 3x^2-4x-4=0 \), so \( x=2\ \text{or}\ -\frac23 \).
2. Angle check: \( \cos\theta=\frac{4x^2-2x+10}{2(10+x^2)} \). For \( x=2 \): \( \frac{22}{28}=\frac{11}{14} \) - works. For \( x=-\frac23 \): \( \frac{118}{188}\ne\frac{11}{14} \) - fails.
3. So \( x=2 \).

Teacher's Note:
1. Use the magnitude condition to get candidates, then test the angle condition.
2. Always check each candidate; a magnitude condition alone is not enough.

Answer: (c) (1) and (3) only

Explanation:
1. Unit direction \( \frac13(2,-1,2) \). Then \( \vec p=\vec a\pm15\cdot\frac13(2,-1,2)=(3,1,-1)\pm5(2,-1,2) \).
2. Plus: \( (13,-4,9) \). Minus: \( (-7,6,-11) \).
3. These are options (1) and (3).

Teacher's Note:
1. A point at distance \( d \) along a line is \( A \) plus or minus \( d \) times the unit direction.
2. Both signs are valid.

Answer: (a) \( -81 \)

Explanation:
1. Normal to the plane of the two given vectors: \( (1,1,-2)\times(1,-2,1)=(-3,-3,-3) \), so \( \vec v\cdot(1,1,1)=0 \), i.e. \( a+b=0 \) (the orthogonality condition gives the same).
2. Then \( \vec v=\lambda a(\hat j-\hat k) \) and the projection length is \( \frac{|\lambda a(-1-1)|}{\sqrt3}=6\sqrt3 \), so \( |\lambda a|=9 \).
3. Therefore \( \lambda^2ab=-\lambda^2a^2=-81 \).

Teacher's Note:
1. Coplanarity with two vectors means perpendicular to their cross product.
2. Use \( b = -a \) to reduce to one unknown.

Answer: (b) (1) and (2) only

Explanation:
1. The vectors \( \hat a,\hat b,\hat a\times\hat b \) are mutually perpendicular unit vectors.
2. A unit vector equally inclined to all three has equal components along them: \( \pm\frac1{\sqrt3}(\hat a+\hat b+\hat a\times\hat b) \) (equal angle with each, taking the signs together).
3. These are options (1) and (2); options (3) and (4) are inclined to \( \hat a\times\hat b \) at the supplementary angle.

Teacher's Note:
1. Treat the three perpendicular unit vectors as an orthonormal basis.
2. Equal inclination means equal direction cosines in absolute value with the same sign pattern.

Answer: (a) (1), (3) and (4) only

Explanation:
1. The coefficients sum to \( 3-2+1-2=0 \), so the four points are coplanar (1) and the four vectors are dependent (4).
2. Rewrite: \( \frac{3\vec a+\vec c}{4}=\frac{\vec b+\vec d}{2} \). The left side is a point on AC with AP : PC = 1 : 3 and the right is the mid-point of BD.
3. So AC cuts BD at its mid-point, ratio 1 : 1 (3). BD divides AC as 1 : 3, so (2) is false.

Teacher's Note:
1. Zero-sum coefficients imply coplanarity.
2. Rearrange to equate two section-formula expressions for the same point.

Answer: (b) \( \vec a+\frac13\frac{|\vec a|}{|\vec c|}\vec c \)

Explanation:
1. \( F \) lies on AB, which is \( \vec a+\lambda\vec c \).
2. \( F \) also lies on line CP: \( \vec c+u(\vec p-\vec c) \) with \( \vec p=\frac{t}{|\vec a|}\vec a+\frac{t}{|\vec c|}\vec c \).
3. The coefficient of a gives \( u\frac{t}{|\vec a|}=1 \); then the coefficient of \( c \) is \( \frac{|\vec a|}{3|\vec c|} \).
4. So \( \vec f=\vec a+\frac13\frac{|\vec a|}{|\vec c|}\vec c \).

Teacher's Note:
1. \( F \) is on both lines, so equate the representations.
2. Use the coefficient of a to fix the parameter.

Answer: (b) \( \frac13\frac{|\vec a|}{|\vec c|}\vec c \)

Explanation:
1. From the position vector of \( F \), \( \overrightarrow{AF}=\vec f-\vec a=\frac13\frac{|\vec a|}{|\vec c|}\vec c \).

Teacher's Note:
1. Subtract the position vector of \( A \).
2. AF is parallel to OC because AB is parallel to OC in the parallelogram.

Answer: (b) \( 1 \)

Explanation:
1. For the reciprocal system, \( \vec a'\times\vec b'=\frac{\vec c}{V} \), \( \vec b'\times\vec c'=\frac{\vec a}{V} \), \( \vec c'\times\vec a'=\frac{\vec b}{V} \).
2. The sum is \( \frac{\vec a+\vec b+\vec c}{V} \), so \( \lambda=1 \).

Teacher's Note:
1. Use the reciprocal-system identity \( a' x b' = c / [a b c] \) and its cyclic forms.
2. Compare with the given right-hand side.

Answer: (a) \( 32 \)

Explanation:
1. The base ABCD is a square of side 4 in the plane \( y=0 \), area 16.
2. The height of \( E \) above this plane is 6.
3. Volume \( =\frac13\times16\times6=32 \).

Teacher's Note:
1. Volume of a pyramid = one third of base area times height.
2. The height is the perpendicular distance, here the y-coordinate of \( E \).

Answer: (c) coplanar and the plane is parallel to the base

Explanation:
1. The four triangular faces are EAB, EBC, ECD, EDA. Each centroid has \( y=\frac{0+0+6}{3}=2 \).
2. So they all lie in the plane \( y=2 \), which is parallel to the base \( y=0 \).

Teacher's Note:
1. The y-coordinate of each centroid is one third of the apex height.
2. A plane \( y \) = constant is parallel to the base plane \( y = 0 \).

Answer: (c) 0

Explanation:
1. A linear system over the real numbers has either no solution, exactly one solution, or infinitely many solutions.
2. Having exactly two distinct solutions is impossible, so the number of such matrices is 0.

Teacher's Note:
1. Think about the structure of solution sets of linear systems.
2. If two solutions exist, every point on the line through them is also a solution.

Answer: (a) \( 3 \)

Explanation:
1. Expand: \( 2\cdot3-2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=9 \), so the sum of dot products is \( -\frac32 \).
2. Then \( |\vec a+\vec b+\vec c|^2=3-3=0 \), so \( \vec b+\vec c=-\vec a \).
3. So \( |2\vec a+5(\vec b+\vec c)|=|2\vec a-5\vec a|=3 \).

Teacher's Note:
1. Expand each squared difference using unit lengths.
2. A zero magnitude means the vector is the zero vector.

Answer: (c) \( 10 \)

Explanation:
1. Since \( \overrightarrow{PQ}+\overrightarrow{PS}=\overrightarrow{PR} \) and \( \overrightarrow{PQ}-\overrightarrow{PS}=\overrightarrow{SQ} \): \( \overrightarrow{PQ}=(2,-1,-3),\ \overrightarrow{PS}=(1,2,1) \).
2. Cross product \( \overrightarrow{PQ}\times\overrightarrow{PS}=(5,-5,5) \).
3. Volume \( =|(1,2,3)\cdot(5,-5,5)|=|5-10+15|=10 \).

Teacher's Note:
1. The diagonals of a parallelogram give its sides by half-sum and half-difference.
2. Volume = |scalar triple product|.

Answer: (a) (2) and (4) only

Explanation:
1. Directions \( (1,2,2) \) and \( (2,2,1) \): \( \ell \) has direction \( (1,2,2)\times(2,2,1)=(-2,3,-2) \), i.e. \( (2,-3,2) \).
2. Intersection with \( \ell_1 \): \( t(2,-3,2) \) gives \( t=-1 \) in \( \ell_1 \), the point \( (2,-3,2) \).
3. Points on \( \ell_2 \): \( 9u^2+28u+37=17 \) gives \( u=-2 \) or \( -\frac{10}9 \), i.e. \( (-1,-1,0) \) and \( \left(\frac79,\frac79,\frac89\right) \).

Teacher's Note:
1. Find the common perpendicular direction by a cross product.
2. Parametrise the second line and use the distance condition.

Answer: (b) \( 4 \)

Explanation:
1. Take \( \vec a=(1,0,0),\ \vec b=\left(\frac12,\frac{\sqrt3}2,0\right),\ \vec c=\left(\frac12,\frac{\sqrt3}6,\frac{\sqrt6}3\right) \) (unit vectors at \( 60^\circ \)).
2. Compute \( \vec a\times\vec b+\vec b\times\vec c \) and solve for the coefficients: \( p=\frac{\sqrt2}2,\ q=-\frac{\sqrt2}2,\ r=\frac{\sqrt2}2 \).
3. Then \( \frac{p^2+2q^2+r^2}{q^2}=\frac{\frac12+1+\frac12}{\frac12}=4 \).

Teacher's Note:
1. Choose convenient coordinates for three unit vectors at 60 degrees.
2. Solve the three linear equations for \( p, q, r \).

Answer: (c) P-3, Q-4, R-1, S-2

Explanation:
1. \( (P) 6[\vec a\times\vec b,\ \vec b\times\vec c,\ \vec c\times\vec a]=6V^2=6\cdot4=24 \).
2. \( (Q) 6[\vec a+\vec b,\ \vec b+\vec c,\ \vec c+\vec a]=6\cdot2V=60 \).
3. \( (R) \frac12|(2\vec a+3\vec b)\times(\vec a-\vec b)|=\frac52|\vec a\times\vec b|=\frac52\cdot40=100 \).
4. \( (S) |(\vec a+\vec b)\times\vec a|=|\vec b\times\vec a|=30 \).

Teacher's Note:
1. Each transformation multiplies the triple product or the cross product by a determinant-type factor.
2. Remember that the triangle area is half the cross product magnitude.

Answer: (a) \( 3 \)

Explanation:
1. Cross product: \( \vec a\times\vec b=(\vec p+2\vec q)\times(2\vec p+\vec q)=\vec p\times\vec q+4\,\vec q\times\vec p=-3\,\vec p\times\vec q \).
2. Magnitude \( 3\sin30^\circ=\frac32 \), which is the area.
3. Twice the area is \( 3 \).

Teacher's Note:
1. Expand the cross product using \( p x p = 0 \) and \( q x p = -p x q \).
2. Area of a parallelogram = \( |a x b| \).

Answer: (d) \( 5 \)

Explanation:
1. Square the sum: \( 9+16+25+2(\vec u\cdot\vec v+\vec v\cdot\vec w+\vec w\cdot\vec u)=0 \).
2. The sum of dot products is \( -25 \), and the square root of its magnitude is 5.

Teacher's Note:
1. Square the vector equation \( u + v + w = 0 \).
2. Take the absolute value before the square root.

Answer: (c) \( -3 \) or \( 4 \)

Explanation:
1. The general solution is \( \vec x=\frac{\beta\vec b+\vec a\times\vec b}{|\vec b|^2} \) (from \( \vec b\times(\vec b\times\vec x)=\vec b\times\vec a \)).
2. Compare the coefficients of \( \vec b \): \( \beta^2-12=\beta \).
3. So \( \beta^2-\beta-12=0 \) and \( \beta=4 \) or \( -3 \).

Teacher's Note:
1. Solve the two vector conditions for \( x \) in general form first.
2. Compare with the given form to get an equation in \( \beta \).

Answer: (d) \( 4 \)

Explanation:
1. Vectors from the first point: \( (4,6,2),\ (3,10,5),\ (-4,5,k+1) \).
2. Determinant \( \begin{vmatrix}4&6&2\\3&10&5\\-4&5&k+1\end{vmatrix}=0 \): \( 4(10k+10-25)-6(3k+3+20)+2(15+40)=0 \).
3. This simplifies to \( 40k-60-18k-138+110=22k-88=0 \), so \( k=4 \).

Teacher's Note:
1. Coplanar points give a zero triple product of three edges.
2. Expand the determinant along the first row.

Answer: (b) \( \frac12 \)

Explanation:
1. Their sum is zero, since \( \overrightarrow{AN}+\overrightarrow{BP}+\overrightarrow{CM}=\vec 0 \) for equal ratios, so they form a triangle.
2. Its area as a fraction of the area of ABC is \( 1-3\lambda+3\lambda^2 \).
3. This is least when \( \lambda=\frac12 \) (value \( \frac14 \)).

Teacher's Note:
1. Write each of the three vectors in terms of AB and AC.
2. Minimise the quadratic in \( \lambda \).

Answer: (a) \( 27 \)

Explanation:
1. The centroid tetrahedron is similar to the original, with ratio \( \frac13 \) (it is the original scaled by \( -\frac13 \) about the centroid).
2. Volumes scale by \( \left(\frac13\right)^3=\frac1{27} \).
3. So \( V=27V' \).

Teacher's Note:
1. The centroid of a face divides the median of the face as 2 : 1.
2. Volume scales as the cube of the linear ratio.

Answer: (b) \( 10\hat i \)

Explanation:
1. The three components must vanish: \( \vec u\cdot\vec R=10,\ \vec v\cdot\vec R=20,\ \vec w\cdot\vec R=20 \).
2. Test \( \vec R=10\hat i \): \( 1\times10=10,\ 2\times10=20,\ 2\times10=20 \) - all satisfied.
3. The coefficient determinant of \( \vec u,\vec v,\vec w \) is \( -5\ne0 \), so the solution is unique: \( \vec R=10\hat i \).

Teacher's Note:
1. Each bracket must be zero because \( i, j, k \) are independent.
2. A non-zero determinant guarantees a unique solution.

Answer: (a) \( -(\vec a\cdot\vec b)\vec a+|\vec a|^2\vec b+\vec a\times\vec b \)

Explanation:
1. Write \( \vec r=x\vec a+y\vec b+z(\vec a\times\vec b) \).
2. From \( [\vec r\,\vec a\,\vec b]=\vec r\cdot(\vec a\times\vec b)=z|\vec a\times\vec b|^2=z=1 \).
3. From \( \vec r\cdot\vec a=0 \): \( x|\vec a|^2+y(\vec a\cdot\vec b)=0 \); from \( \vec r\cdot\vec b=1 \): \( x(\vec a\cdot\vec b)+y|\vec b|^2=1 \).
4. Solving, \( x=-(\vec a\cdot\vec b),\ y=|\vec a|^2 \).

Teacher's Note:
1. Expand \( r \) in the basis a, \( b, a x b \).
2. Use the three given conditions to find the three coefficients.

Answer: (c) \( \frac{p^2}{a^2+b^2+c^2} \)

Explanation:
1. By the Cauchy-Schwarz inequality: \( (ax+by+cz)^2\le(a^2+b^2+c^2)(x^2+y^2+z^2) \).
2. So \( x^2+y^2+z^2\ge\frac{p^2}{a^2+b^2+c^2} \), with equality when \( (x,y,z) \) is parallel to \( (a,b,c) \).

Teacher's Note:
1. This is the squared distance from the origin to the plane.
2. Equality in Cauchy-Schwarz holds for proportional triples.