JEE Mathematics Vector Algebra MCQs Set 08

Answer: (c) \( \vec r=x\vec a+\dfrac{\vec a\times\vec b}{|\vec a|^2} \)

Explanation:
1. Take \( \vec r=x\vec a+\frac{\vec a\times\vec b}{|\vec a|^2} \). Then \( \vec r\times\vec a=\frac{(\vec a\times\vec b)\times\vec a}{|\vec a|^2} \), since \( \vec a\times\vec a=\vec 0 \).
2. \( (\vec a\times\vec b)\times\vec a=\vec b(\vec a\cdot\vec a)-\vec a(\vec a\cdot\vec b)=|\vec a|^2\vec b \) because \( \vec a\cdot\vec b=0 \).
3. So \( \vec r\times\vec a=\vec b \), and the term \( x\vec a \) is the free part.

Teacher's Note:
1. The term x a solves the homogeneous equation r x a = 0, so only a particular solution is needed.

Answer: (b) 7

Explanation:
1. Take the first vertex as \(A\): \( \vec{AB}=(-2,3,-3),\ \vec{AC}=(4,5,\lambda-10),\ \vec{AD}=(6,2,-3) \).
2. The scalar triple product is \( 22(\lambda-4) \). The volume is \( \frac16|22(\lambda-4)|=11 \), so \( |\lambda-4|=3 \).
3. Hence \( \lambda=7 \) or \( \lambda=1 \); only 7 is among the options.

Teacher's Note:
1. Volume of a tetrahedron is one sixth of the absolute scalar triple product of three edges from one vertex.

Answer: (b) \( x=1 \)

Explanation:
1. \( \vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c \). Comparing coefficients (b and c independent): \( \vec a\cdot\vec c=x^2-2x+6 \) and \( -\vec a\cdot\vec b=\sin y \).
2. From \( \vec a\cdot\vec b+\vec a\cdot\vec c=4 \): \( x^2-2x+6-\sin y=4 \), i.e. \( (x-1)^2+1=\sin y \).
3. Since \( \sin y\le1 \), we need \( x=1 \) and \( \sin y=1 \). So the point lies on \( x=1 \).

Teacher's Note:
1. The left side is at least 1 and the right side at most 1, so both must equal 1.

Answer: (a) \( \hat i \)

Explanation:
1. \( \vec a\times(\vec a\times\vec b)=\vec a(\vec a\cdot\vec b)-\vec b|\vec a|^2 \). The left side is \( (1,1,1)\times(0,1,-1)=(-2,1,1) \).
2. So \( (1,1,1)-3\vec b=(-2,1,1) \), giving \( \vec b=(1,0,0) \).
3. Check: \( \vec a\cdot\hat i=1 \) and \( \vec a\times\hat i=\hat j-\hat k \). So \( \vec b=\hat i \).

Teacher's Note:
1. Cross a with the given cross product to recover b.

Answer: (d) 9

Explanation:
1. The expression equals \( 6-2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a) \).
2. Since \( |\vec a+\vec b+\vec c|^2=3+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)\ge0 \), the sum of dot products is at least \( -\frac32 \).
3. So the expression is at most \( 6+3=9 \).

Teacher's Note:
1. Use |a + b + c|^2 >= 0 to bound the sum of dot products.

Answer: (b) \( \frac{2\sqrt2}3 \)

Explanation:
1. \( (\vec a\times\vec b)\times\vec c=\vec b(\vec a\cdot\vec c)-\vec a(\vec b\cdot\vec c) \). Setting this equal to \( \frac13|\vec b||\vec c|\vec a \): \( \vec b(\vec a\cdot\vec c)=\vec a\left(\vec b\cdot\vec c+\frac13|\vec b||\vec c|\right) \).
2. Taking \( \vec a,\vec b \) non-parallel, both coefficients vanish: \( \vec b\cdot\vec c=-\frac13|\vec b||\vec c| \), i.e. \( \cos\theta=-\frac13 \).
3. Then \( \sin\theta=\sqrt{1-\frac19}=\frac{2\sqrt2}3 \).

Teacher's Note:
1. Expand the vector triple product and use that a and b are independent.

Answer: (c) \( V^3 \)

Explanation:
1. The coefficients of \( \vec\alpha,\vec\beta,\vec\gamma \) in terms of \( \vec a,\vec b,\vec c \) form the Gram matrix of \( \vec a,\vec b,\vec c \).
2. Hence \( [\vec\alpha\,\vec\beta\,\vec\gamma]=\det(\text{Gram})\cdot[\vec a\,\vec b\,\vec c] \).
3. \( \det(\text{Gram})=[\vec a\,\vec b\,\vec c]^2=V^2 \), so the new volume is \( V^3 \).

Teacher's Note:
1. The determinant of the Gram matrix of three vectors equals the square of their scalar triple product.

Answer: (b) \( p=1 \) or \( p=-\frac13 \)

Explanation:
1. Rotating the axes does not change the magnitude of the vector.
2. So \( (2p)^2+1^2=(p+1)^2+1^2 \), i.e. \( 4p^2=(p+1)^2 \).
3. \( 3p^2-2p-1=0 \), so \( p=1 \) or \( p=-\frac13 \).

Teacher's Note:
1. A rotation of the axes preserves length.

Answer: (a) 4

Explanation:
1. Take \( A \) as the origin with \( \vec{AB}=\vec b,\ \vec{AC}=\vec c,\ \vec{AD}=\vec d \).
2. The sum is \( \vec b\times(\vec d-\vec c)+(\vec c-\vec b)\times\vec d+(-\vec c)\times(\vec d-\vec b) \), which simplifies to \( -2\,\vec b\times\vec c \).
3. Its magnitude is \( 2|\vec b\times\vec c|=2\cdot2\Delta=4\Delta \).

Teacher's Note:
1. Choose A as the origin and expand; all the terms with d cancel.

Answer: (a) \( \frac34\sqrt{10} \)

Explanation:
1. \( AB=\sqrt6 \) and \( AC=\sqrt{54}=3\sqrt6 \), so the bisector meets \( BC \) at \(D\) with \( BD:DC=1:3 \).
2. \( D=\frac{3B+C}4=\left(\frac14,\frac54,-3\right) \).
3. \( \vec{AD}=\left(-\frac34,\frac94,0\right) \), so \( AD=\sqrt{\frac{9+81}{16}}=\frac{3\sqrt{10}}4 \).

Teacher's Note:
1. The angle bisector divides the opposite side in the ratio of the adjacent sides.

Answer: (a) \( \frac12 \)

Explanation:
1. The determinant is \( [\vec a\,\vec b\,\vec c] \). Because \( \vec c \) is perpendicular to \( \vec a \) and \( \vec b \), \( \vec c \) is parallel to \( \vec a\times\vec b \), so \( |[\vec a\,\vec b\,\vec c]|=|\vec a\times\vec b||\vec c|=|\vec a||\vec b|\sin\frac\pi4 \).
2. Squaring: \( [\vec a\,\vec b\,\vec c]^2=|\vec a|^2|\vec b|^2\cdot\frac12 \).
3. So \( \lambda=\frac12 \).

Teacher's Note:
1. c is parallel to a x b, so the triple product is |a x b| times |c|.

Answer: (c) form an equilateral triangle

Explanation:
1. The differences of the position vectors are \( (\alpha-\beta,\ \beta-\gamma,\ \gamma-\alpha) \), \( (\beta-\gamma,\ \gamma-\alpha,\ \alpha-\beta) \) and \( (\gamma-\alpha,\ \alpha-\beta,\ \beta-\gamma) \), each a cyclic rearrangement of the others.
2. All three sides have length \( \sqrt{(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2} \), which is non-zero.
3. So the triangle is equilateral.

Teacher's Note:
1. Compute the three side vectors; they are cyclic shifts of each other.

Answer: (c) \( 2|\vec r|^2 \)

Explanation:
1. Let \( \vec r=x\hat i+y\hat j+z\hat k \). Then \( |\vec r\times\hat i|^2=y^2+z^2 \), \( |\vec r\times\hat j|^2=x^2+z^2 \) and \( |\vec r\times\hat k|^2=x^2+y^2 \).
2. The sum is \( 2\left(x^2+y^2+z^2\right) \).
3. So the value is \( 2|\vec r|^2 \).

Teacher's Note:
1. |r x i|^2 is the squared distance of r from the x-axis.

Answer: (b) \( \sin\frac\theta2 \)

Explanation:
1. \( |\vec a-\vec b|^2=1+1-2\cos\theta=2(1-\cos\theta)=4\sin^2\frac\theta2 \).
2. So \( |\vec a-\vec b|=2\sin\frac\theta2 \).
3. Dividing by 2, the value is \( \sin\frac\theta2 \).

Teacher's Note:
1. Use 1 - cos(theta) = 2 sin^2(theta/2).

Answer: (a) \( \hat i-\hat j-5\hat k \)

Explanation:
1. The moment of the couple is \( \vec r\times\vec F \), where \( \vec r \) joins the point of application of \( -\vec F \) to that of \( \vec F \): \( \vec r=(9-3,\,-1+2,\,2-1)=(6,1,1) \).
2. \( (6,1,1)\times(5,0,1)=(1\cdot1-1\cdot0,\ 1\cdot5-6\cdot1,\ 6\cdot0-1\cdot5)=(1,-1,-5) \).
3. So the moment is \( \hat i-\hat j-5\hat k \).

Teacher's Note:
1. The moment of a couple is the vector joining the two points crossed with either force.

Answer: (b) G.M. of \(a\) and \(b\)

Explanation:
1. With \( \hat x,\hat y,\hat x\times\hat y \) as an orthonormal basis, the vectors are \( (a,a,c),\ (1,0,1),\ (c,c,b) \).
2. Coplanarity requires \( \begin{vmatrix}a&a&c\\1&0&1\\c&c&b\end{vmatrix}=0 \), which expands to \( c^2-ab=0 \).
3. So \( c^2=ab \) and \(c\) is the geometric mean of \(a\) and \(b\).

Teacher's Note:
1. Use the right-handed orthonormal triad x, y, x cross y as a coordinate system.

Answer: (c) \( 2|\vec a|^2 \)

Explanation:
1. With \( \vec a=(x,y,z) \): \( |\vec a\times\hat i|^2=y^2+z^2 \), \( |\vec a\times\hat j|^2=x^2+z^2 \), \( |\vec a\times\hat k|^2=x^2+y^2 \).
2. The sum is \( 2(x^2+y^2+z^2) \).
3. So the value is \( 2|\vec a|^2 \).

Teacher's Note:
1. Each term is the squared distance of the point from one coordinate axis.

Answer: (b) \( -4\vec c \)

Explanation:
1. The first line is \( -4\vec c+t(6\vec a-4\vec b-\vec c) \) and the second is \( (-1+2s)\vec a+(-2+4s)\vec b+(-3-2s)\vec c \).
2. Matching coefficients (a, b, c independent): \( 6t=-1+2s \), \( -4t=-2+4s \), \( -4-t=-3-2s \). The first two give \( t=0,\ s=\frac12 \), and the third is satisfied.
3. The point is \( -4\vec c \).

Teacher's Note:
1. Treat a, b, c as an independent basis and compare coefficients.

Answer: (d) \( \pm\sqrt3(\hat i+\hat j+\hat k) \)

Explanation:
1. The equal components give \( \vec r=t(\hat i+\hat j+\hat k) \).
2. \( |\vec r|=\sqrt3|t|=3 \), so \( t=\pm\sqrt3 \).
3. So \( \vec r=\pm\sqrt3(\hat i+\hat j+\hat k) \).

Teacher's Note:
1. Equal dot products with i, j, k mean equal coordinates.

Answer: (a) \( -74 \)

Explanation:
1. \( (\vec\alpha\times\vec\beta)\cdot(\vec\alpha\times\vec\gamma)=(\vec\alpha\cdot\vec\alpha)(\vec\beta\cdot\vec\gamma)-(\vec\alpha\cdot\vec\gamma)(\vec\alpha\cdot\vec\beta) \).
2. \( \vec\alpha\cdot\vec\alpha=14,\ \vec\beta\cdot\vec\gamma=-3,\ \vec\alpha\cdot\vec\gamma=4,\ \vec\alpha\cdot\vec\beta=-2+6+4=8 \).
3. The value is \( 14(-3)-4(8)=-74 \).

Teacher's Note:
1. The dot product of two cross products has a standard determinant expansion.

Answer: (c) \( \vec{DC} \)

Explanation:
1. \( \vec{BE}=\frac{\vec A+\vec C}2-\vec B \) and \( \vec{AF}=\frac{\vec B+\vec C}2-\vec A \).
2. Adding: \( \vec C-\frac{\vec A+\vec B}2 \).
3. This is \( \vec{DC} \), since \( D \) is the midpoint of \( AB \).

Teacher's Note:
1. Write every vector as a difference of position vectors.

Answer: (a) \( \vec{PA}+\vec{PB}=2\vec{PC} \)

Explanation:
1. With \(P\) as the origin, the position vector of \(C\) is \( \frac{\vec{PA}+\vec{PB}}2 \).
2. So \( \vec{PA}+\vec{PB}=2\vec{PC} \).
3. Hence option b.

Teacher's Note:
1. The midpoint is the average of the endpoints measured from any common origin.

Answer: (d) \( \cos^{-1}\left(\frac1{\sqrt3}\right) \)

Explanation:
1. Let the common magnitude be \(m\). Then \( \vec a\cdot(\vec a+\vec b+\vec c)=m^2 \) and \( |\vec a+\vec b+\vec c|=\sqrt{3}\,m \).
2. \( \cos\theta=\frac{m^2}{m\cdot\sqrt3m}=\frac1{\sqrt3} \).
3. So \( \theta=\cos^{-1}\frac1{\sqrt3} \).

Teacher's Note:
1. The cross terms vanish because the vectors are mutually perpendicular.

Answer: (b) \( \vec r=-\vec q+\frac{(\vec p\cdot\vec q)}{(\vec p\cdot\vec p)}\vec p \)

Explanation:
1. The foot \(F\) of the perpendicular from \(B\) on line \(AD\) satisfies \( \vec{AF}=\frac{\vec p\cdot\vec q}{\vec p\cdot\vec p}\vec p \) (projection of \( \vec{AB} \) on \( \vec{AD} \)).
2. The altitude vector from \(B\) to \(F\) is \( \vec{BF}=\vec{AF}-\vec{AB} \).
3. So \( \vec r=-\vec q+\frac{\vec p\cdot\vec q}{\vec p\cdot\vec p}\vec p \).

Teacher's Note:
1. Subtract the projection of q on p from q, then reverse the direction to run from B to the foot.

Answer: (b) 3

Explanation:
1. \( \vec{DB}=\vec{AB}-\vec{AD} \), so \( \vec{DB}\cdot\vec{AB}=a^2-\vec{AD}\cdot\vec{AB} \).
2. From \( \vec{AC}=\vec{AB}+\vec{AD} \): \( c^2=a^2+b^2+2\vec{AB}\cdot\vec{AD} \), so \( \vec{AB}\cdot\vec{AD}=\frac{c^2-a^2-b^2}2 \).
3. Therefore \( \vec{DB}\cdot\vec{AB}=a^2-\frac{c^2-a^2-b^2}2=\frac{3a^2+b^2-c^2}2 \) and \( k=3 \).

Teacher's Note:
1. Express both vectors through AB and AD, then use the law of cosines for the diagonal.

Answer: (c) 3

Explanation:
1. The expression is \( 6+2(\hat x\cdot\hat y+\hat y\cdot\hat z+\hat z\cdot\hat x) \).
2. Since \( |\hat x+\hat y+\hat z|^2=3+2(\hat x\cdot\hat y+\hat y\cdot\hat z+\hat z\cdot\hat x)\ge0 \), the sum of the dot products is at least \( -\frac32 \).
3. The minimum is \( 6-3=3 \), attained when \( \hat x+\hat y+\hat z=\vec0 \).

Teacher's Note:
1. Use the same non-negativity trick as for the sum of squared differences.

Answer: (b) 7

Explanation:
1. \( \vec x\times\vec y=(18+4,\ -1+9,\ 12+6)=(22,8,18) \).
2. \( |\vec z|=\sqrt{9+16+144}=13 \) and \( (\vec x\times\vec y)\cdot\vec z=66-32-216=-182 \).
3. The magnitude of the projection is \( \frac{182}{13}=14=2k \), so \( k=7 \).

Teacher's Note:
1. Projection of u on z is (u . z)/|z|.

Answer: (b) 3

Explanation:
1. \( \hat a+\hat c=\sqrt3\hat b \), so \( |\hat a+\hat c|^2=3 \).
2. \( 1+1+2\hat a\cdot\hat c=3 \), so \( \hat a\cdot\hat c=\frac12 \).
3. The angle is \( \frac\pi3 \), so \( k=3 \).

Teacher's Note:
1. Square the relation a + c = sqrt(3) b.

Answer: (d) 5

Explanation:
1. Collinear vectors have proportional coefficients: \( \frac{\alpha-2}{2+3\alpha}=\frac1{-3} \).
2. \( -3\alpha+6=2+3\alpha \), so \( 6\alpha=4 \) and \( \alpha=\frac23 \).
3. Then \( l=2,\ m=3 \) and \( l+m=5 \).

Teacher's Note:
1. Because a and b are independent, compare coefficients of a and b.

Answer: (c) 3

Explanation:
1. \( \vec c\cdot\vec d=5|\hat a|^2-4\hat a\cdot\hat b+10\hat a\cdot\hat b-8|\hat b|^2=-3+6\hat a\cdot\hat b=0 \).
2. So \( \hat a\cdot\hat b=\frac12 \), i.e. the angle is \( \frac\pi3 \).
3. Hence \( k=3 \).

Teacher's Note:
1. Expand the dot product and use |a| = |b| = 1.