JEE Mathematics Vectors MCQs Set 30

Answer: (d) \( \frac32 \)

Explanation:
1. \( |\vec c-\vec a|^2=|\vec c|^2-2\vec a\cdot\vec c+|\vec a|^2 \) gives \( 8=|\vec c|^2-2|\vec c|+9 \), so \( (|\vec c|-1)^2=0 \) and \( |\vec c|=1 \).
2. \( \vec a\times\vec b=(2,-2,1) \), so \( |\vec a\times\vec b|=3 \).
3. \( |(\vec a\times\vec b)\times\vec c|=3\cdot1\cdot\sin30^\circ=\dfrac32 \).

Teacher's Note:
1. Use the magnitude of a cross product: \( |\vec p\times\vec q|=|\vec p||\vec q|\sin\theta \).
2. Solve for \( |\vec c| \) from the given conditions first.

Answer: (b) \( 3 \)

Explanation:
1. The sum equals \( 6+2(\hat x\cdot\hat y+\hat y\cdot\hat z+\hat z\cdot\hat x) \).
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 bracket is at least \( -\tfrac32 \).
3. Minimum \( =6-3=3 \), attained when \( \hat x+\hat y+\hat z=\vec 0 \).

Teacher's Note:
1. Use the non-negativity of the squared magnitude of the sum.
2. Equality needs the three unit vectors to sum to zero (e.g. coplanar at 120 degrees).

Answer: (b) \( \theta\in[0,2\pi/3] \)

Explanation:
1. \( \cos\theta=\dfrac{ab+bc+ca}{a^2+b^2+c^2} \).
2. The numerator satisfies \( ab+bc+ca\le a^2+b^2+c^2 \) and \( ab+bc+ca\ge-\tfrac12(a^2+b^2+c^2) \) (from \( (a+b+c)^2\ge0 \)).
3. So \( -\tfrac12\le\cos\theta\le1 \), i.e. \( \theta\in[0,2\pi/3] \).

Teacher's Note:
1. Both bounds on \( ab+bc+ca \) come from squares: \( (a-b)^2+(b-c)^2+(c-a)^2\ge0 \) and \( (a+b+c)^2\ge0 \).
2. Equality \( \cos\theta=-\tfrac12 \) occurs when \( a+b+c=0 \).

Answer: (d) never coplanar

Explanation:
1. With \( x=\cos\alpha,\ y=\cos\beta,\ z=\cos\gamma \), the scalar triple product is \( \begin{vmatrix}x&1&1\\1&y&1\\1&1&z\end{vmatrix}=xyz-x-y-z+2 \).
2. This expression is linear in each of \( x,y,z \) on \( [-1,1] \), so its minimum is at the end-points; there it is \( \ge0 \), and equals 0 only when at least two of the values are 1.
3. Since none of the cosines can equal 1, the expression is strictly positive: the vectors are never coplanar.

Teacher's Note:
1. Three vectors are coplanar exactly when their scalar triple product is zero.
2. A function linear in each variable attains its extremes at the corners of the box.

Answer: (a) i, iii and iv only

Explanation:
1. \( \vec a+\vec b+\vec c=\vec 0 \). So \( \vec b\cdot\vec c=-\vec b\cdot\vec a-|\vec b|^2=24 \), giving \( \vec a\cdot\vec b=-72 \) (iv true).
2. \( |\vec c|^2=144+48+2(-72)=48 \), so \( \dfrac{|\vec c|^2}2-|\vec a|=24-12=12 \) (i true) and \( \dfrac{|\vec c|^2}2+|\vec a|=36\ne30 \) (ii false).
3. \( \vec c\times\vec a=-(\vec a+\vec b)\times\vec a=\vec a\times\vec b \), so the expression is \( 2|\vec a\times\vec b| \). \( |\vec a\times\vec b|^2=144\cdot48-72^2=1728 \), so \( |\vec a\times\vec b|=24\sqrt3 \) and the value is \( 48\sqrt3 \) (iii true).

Teacher's Note:
1. Use \( \vec a+\vec b+\vec c=\vec0 \) to express everything in two vectors.
2. Lagrange's identity: \( |\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2 \).

Answer: (d) All of (i), (ii), (iii) and (iv)

Explanation:
1. Each pair satisfies \( \vec x\cdot\vec y=\sqrt2\sqrt2\cos60^\circ=1 \). Then \( \vec a\parallel\vec x\times(\vec y\times\vec z)=\vec y(\vec x\cdot\vec z)-\vec z(\vec x\cdot\vec y)=\vec y-\vec z \).
2. Writing \( \vec a=\lambda(\vec y-\vec z) \): \( \vec a\cdot\vec y=\lambda(2-1)=\lambda \), so \( \vec a=(\vec a\cdot\vec y)(\vec y-\vec z) \) (ii) and equivalently \( \vec a=-(\vec a\cdot\vec y)(\vec z-\vec y) \) (iv).
3. Similarly \( \vec b\parallel\vec y\times(\vec z\times\vec x)=\vec z-\vec x \) with \( \vec b\cdot\vec z=\mu \) giving (i), and \( \vec a\cdot\vec b=\lambda\mu(\vec y-\vec z)\cdot(\vec z-\vec x)=\lambda\mu(1-1-2+1)=-\lambda\mu \) (iii).

Teacher's Note:
1. A vector perpendicular to both \( \vec x \) and \( \vec y\times\vec z \) lies along \( \vec x\times(\vec y\times\vec z) \); expand with the triple product formula.
2. All four statements are equivalent forms once the direction is known.

Answer: (d) 4

Explanation:
1. \( B \) is fixed at \( (d,0,0) \).
2. \( C \) lies in the \( xy \)-plane with \( AC=BC=d \): two positions, above or below the \( x \)-axis.
3. For each triangle \( ABC \), \( D \) can be on either side of the plane: 2 positions. Total \( 2\times2=4 \).

Teacher's Note:
1. Count independent choices: side of the line for C and side of the plane for D.
2. The apex of a regular tetrahedron lies on the perpendicular through the centroid of the base.

Answer: (b) 2

Explanation:
1. Dot with \( \vec u \): \( \vec w\cdot\vec u=\vec v\cdot\vec u \). Taking squared magnitudes: \( 1=|\vec w|^2+|\vec w\times\vec u|^2=2|\vec w|^2-(\vec w\cdot\vec u)^2 \).
2. \( \vec u\times\vec v=\vec u\times\vec w+\vec w-(\vec u\cdot\vec w)\vec u \), so \( (\vec u\times\vec v)\cdot\vec w=|\vec w|^2-(\vec u\cdot\vec w)^2=1-|\vec w|^2 \).
3. Since \( (\vec u\cdot\vec w)^2\le|\vec w|^2 \), \( |\vec w|^2\le1 \); also \( |\vec w|^2\ge\tfrac12 \). So the triple product is at most \( \tfrac12 \) and the least value of the inverse is \( 2 \).

Teacher's Note:
1. Express the scalar triple product in terms of \( |\vec w|^2 \) only.
2. The reciprocal is smallest when the triple product is largest.

Answer: (c) 5

Explanation:
1. Perpendicular means \( x_1+x_2+x_3=0 \).
2. Counting integer solutions with each \( x_i\in\{-3,\dots,2\} \), i.e. shifting by 3 to \( y_i=x_i+3\in\{0,\dots,5\} \) with \( y_1+y_2+y_3=9 \): total solutions \( \binom{11}{2}=55 \) minus those with some \( y_i\ge6 \): \( 3\binom52=30 \), giving \( 25 \).
3. So \( 5k=25 \) and \( k=5 \).

Teacher's Note:
1. Convert to a bounded stars-and-bars count.
2. Check the arithmetic with inclusion-exclusion: only one variable can exceed 5 at a time here.

Answer: (c) 5

Explanation:
1. \( P=(-1,\,2^{1})=(-1,2) \) and \( Q=(2,\,2^{4})=(2,16) \).
2. \( \overrightarrow{OQ}-4\overrightarrow{OP}=(2+4,\,16-8)=(6,8) \), with magnitude \( 10 \).
3. \( 2k=10 \), so \( k=5 \).

Teacher's Note:
1. The condition \( \overrightarrow{OP}\cdot\hat i=-1 \) just says the x-coordinate of P is -1.
2. Substitute the x-coordinates into the curve to get the y-coordinates.

Answer: (c) 4

Explanation:
1. Dot both sides with \( \vec a,\vec b,\vec c \) in turn; with \( \vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=\tfrac12 \) and the triple product \( [\vec a\,\vec b\,\vec c]=V \):
2. \( \vec a:\ \vec b\times\vec c\cdot\vec a=V=p+\tfrac q2+\tfrac r2 \); \( \vec b:\ 0=\tfrac p2+q+\tfrac r2 \); \( \vec c:\ \vec a\times\vec b\cdot\vec c=V=\tfrac p2+\tfrac q2+r \).
3. Subtracting the first and third gives \( p=r \); the second gives \( p+q... \) i.e. \( p+r=-2q \), so \( p=r=-q \). Then \( \dfrac{p^2+2q^2+r^2}{q^2}=\dfrac{q^2+2q^2+q^2}{q^2}=4 \).

Teacher's Note:
1. Use dot products to turn the vector equation into three scalar equations.
2. The triple product terms cancel when the first and third equations are subtracted.