JEE Mathematics Vectors MCQs Set 22

Answer: (a) \( 16\vec b \)

Explanation:
1. Use \( \vec a\times(\vec a\times\vec v)=(\vec a\cdot\vec v)\vec a-|\vec a|^2\vec v \). For \( \vec v=\vec b \) (with a.b = 0): \( \vec a\times(\vec a\times\vec b)=-4\vec b \).
2. Then \( \vec a\times(\vec a\times(\vec a\times\vec b))=-4(\vec a\times\vec b) \).
3. One more step: \( \vec a\times(-4\,\vec a\times\vec b)=-4(-4\vec b)=16\vec b \).

Teacher's Note:
1. Expand a × (a × v) with the vector triple product formula.
2. Each double cross product multiplies by −|a|² = −4.

Answer: (b) \( \vec a \) is perpendicular to \( \vec b \)

Explanation:
1. From the given relation \( 3\vec a+4\vec b=-5\vec c \). Squaring: \( 9+16+24\,\vec a\cdot\vec b=25 \).
2. So \( \vec a\cdot\vec b=0 \): a is perpendicular to b.

Teacher's Note:
1. Square both sides of 3a + 4b = −5c.
2. Use |a| = |b| = |c| = 1.

Answer: (b) 9

Explanation:
1. Expanding: \( =6-2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a) \).
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 \( 6+3=9 \) (attained when a + b + c = 0).

Teacher's Note:
1. Expand each square using unit vectors.
2. Use |a + b + c|² ≥ 0.

Answer: (b) \( -\frac{\sqrt3}2,\ \frac{\sqrt3}2 \)

Explanation:
1. \( (\hat i\times\hat j)\cdot\vec c=\hat k\cdot\vec c=c_z \), the z-component of \( \vec c \).
2. Write \( \vec c=\cos\frac\pi3\,\hat u+\sin\frac\pi3\,\hat w \) with \( \hat u=\frac{\hat i+\hat j}{\sqrt2} \) and a unit vector \( \hat w\perp\hat u \). Then \( c_z=\frac{\sqrt3}2\,w_z \).
3. Because \( \hat k\perp\hat u \), \( w_z \) ranges over [−1, 1], so \( c_z\in\left[-\frac{\sqrt3}2,\frac{\sqrt3}2\right] \).

Teacher's Note:
1. The scalar triple product reduces to the z-component of c.
2. Resolve c along and perpendicular to i + j.

Answer: (c) 5

Explanation:
1. \( |\vec a+\vec b|^2=9+16+2\vec a\cdot\vec b=25 \), so \( \vec a\cdot\vec b=0 \).
2. Hence \( |\vec a-\vec b|^2=9+16-0=25 \) and \( |\vec a-\vec b|=5 \).

Teacher's Note:
1. Find a·b from |a + b|.
2. Use it in |a − b|².

Answer: (d) \( \alpha=\pm1,\ \beta=1 \)

Explanation:
1. Linear dependence: \( \begin{vmatrix}1&1&1\\4&3&4\\1&\alpha&\beta\end{vmatrix}=0 \), i.e. \( (3\beta-4\alpha)-(4\beta-4)+(4\alpha-3)=1-\beta=0 \), so \( \beta=1 \).
2. |c|² = 3: \( 1+\alpha^2+\beta^2=3 \), so \( \alpha^2=1 \) and \( \alpha=\pm1 \).

Teacher's Note:
1. Set the determinant of the three vectors to zero.
2. Then use |c|² = 3.

Answer: (c) \( \vec 0 \)

Explanation:
1. Combine the terms: \( \vec r\times\left[(\vec r\cdot\hat i)\hat i+(\vec r\cdot\hat j)\hat j+(\vec r\cdot\hat k)\hat k\right] \).
2. The bracket equals \( \vec r \), so the sum is \( \vec r\times\vec r=\vec 0 \).

Teacher's Note:
1. Take the cross product with r outside.
2. Σ (r·i) i = r.

Answer: (b) \( \frac32 \)

Explanation:
1. |a| = 3. From \( |\vec c-\vec a|^2=8 \): \( |\vec c|^2-2\vec a\cdot\vec c+9=8 \), i.e. \( |\vec c|^2-2|\vec c|+1=0 \), so \( |\vec c|=1 \).
2. a × b = (2, −2, 1) has magnitude 3.
3. So \( |(\vec a\times\vec b)\times\vec c|=3\cdot1\cdot\sin30^\circ=\frac32 \).

Teacher's Note:
1. Use a·c = |c| in |c − a|² to find |c|.
2. Magnitude of a cross product = product of magnitudes × sin of the angle.

Answer: (d) −4

Explanation:
1. From \( \vec a\times(\vec b-2\vec c)=\vec0 \) the vector \( \vec b-2\vec c \) is parallel to \( \vec a \): \( \vec b-2\vec c=\lambda\vec a \).
2. From \( |\vec b\times\vec c|=4\sin\phi=\sqrt{15} \): \( \sin\phi=\frac{\sqrt{15}}4 \), so \( \vec b\cdot\vec c=\pm1 \).
3. Then \( \lambda^2=|\vec b-2\vec c|^2=16+4-4\,\vec b\cdot\vec c \), which is 16 for b·c = 1 and 24 for b·c = −1. So \( \lambda=\pm4 \) (or \( \pm2\sqrt6 \)), and the only option among these is −4.

Teacher's Note:
1. a × (b − 2c) = 0 means b − 2c ∥ a.
2. Take the squared magnitude and use b·c = ±1.

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

Explanation:
1. Points on the first line: \( (6-6s)\vec a+(-4+4s)\vec b+(-5+s)\vec c \). On the second: \( (-1+2t)\vec a+(-2+4t)\vec b+(-3-2t)\vec c \).
2. For non-coplanar a, b, c equate the coefficients: \( 6-6s=-1+2t \) and \( -4+4s=-2+4t \). The second gives \( t=s-\frac12 \); then s = 1, t = 1/2.
3. Check the c-coefficients: −5 + 1 = −4 = −3 − 1. The point is \( -4\vec c \).

Teacher's Note:
1. Parametrise both lines.
2. Equate coefficients (a, b, c are non-coplanar).

Answer: (d) none of these

Explanation:
1. Since \( |\vec b|=|\vec c| \), equal angles means \( \vec a\cdot\vec b=\vec a\cdot\vec c \): \( xy-2yz+3xz=2xz+3xy-yz \), i.e. \( xz-2xy-yz=0 \). Perpendicular to d: \( x-y+2z=0 \).
2. Solving these with \( x^2+y^2+z^2=4 \) and \( y<0 \) gives \( \vec a=\frac2{\sqrt3}(1,-1,-1) \).
3. This is not among the options (2, −2, −2 has magnitude \( 2\sqrt3 \)), so the answer is none of these.

Teacher's Note:
1. Note that |b| = |c|, so equal angles means equal dot products.
2. Check the magnitude of each option.

Answer: (b) \( 9\hat i+5\hat j-2\hat k \)

Explanation:
1. Let \( A=(9,5,5) \) and \( \vec d=(1,3,5) \). Then \( \vec{AP}=(-2,-6,-3) \) and \( \vec{AP}\cdot\vec d=-2-18-15=-35 \), \( |\vec d|^2=35 \).
2. The foot of the perpendicular is \( A-\vec d=(8,2,0) \).
3. The image is \( 2(8,2,0)-(7,-1,2)=(9,5,-2) \), i.e. \( 9\hat i+5\hat j-2\hat k \).

Teacher's Note:
1. Find the foot of the perpendicular from P to the line.
2. The foot is the midpoint of P and its image.

Answer: (a) \( \frac{19}{5\sqrt{43}} \)

Explanation:
1. Let \( p=|\vec a|^2,\ q=|\vec b|^2,\ s=\vec a\cdot\vec b \). From the first pair: \( 6p-7s-5q=0 \). From the second: \( s-p+4q-4s=0 \), i.e. \( -3s-p+4q=0 \).
2. Solving: \( s=\frac{19}{25}q \) and \( p=\frac{43}{25}q \).
3. Then \( \cos\theta=\frac{s}{\sqrt{pq}}=\frac{19/25}{\sqrt{43}/5}=\frac{19}{5\sqrt{43}} \).

Teacher's Note:
1. Expand each dot product in terms of |a|², |b|², a·b.
2. Solve the two equations in terms of |b|².

Answer: (b) \( (-\infty,-6]\cup[-\tfrac16,\infty) \)

Explanation:
1. The initial point is \( \left(\frac5{k+1},\frac{5k}{k+1}\right) \) (k ≠ −1), so \( |\vec b|^2=\frac{25(1+k^2)}{(k+1)^2}\le37 \).
2. Then \( 25+25k^2\le37k^2+74k+37 \), i.e. \( 6k^2+37k+6\ge0 \), or \( (6k+1)(k+6)\ge0 \).
3. So \( k\le-6 \) or \( k\ge-\frac16 \) (k = −1 lies in between and is automatically excluded).

Teacher's Note:
1. Use the section formula for the initial point.
2. Solve the quadratic inequality.

Answer: (d) 3/4

Explanation:
1. \( \vec a\times(\vec a\times\vec b)=(\vec a\cdot\vec b)\vec a-\vec b \), whose magnitude equals \( |\vec a||\vec a\times\vec b|=5\sin\theta \) (it is perpendicular to a).
2. So \( 5\sin\theta=3 \), \( \sin\theta=\frac35 \) and \( \cos\theta=\pm\frac45 \).
3. Hence \( \tan\theta=\pm\frac34 \), and the option available is 3/4.

Teacher's Note:
1. a × (a × b) = (a·b)a − b has magnitude |a||a × b|.
2. Find sin θ from |c| = 3.

Answer: (b) 1 or −1

Explanation:
1. The coefficient determinant is \( \begin{vmatrix}1&1&1\\1&1&0\\0&1&1\end{vmatrix}=1(1)-1(1)+1(1)=1 \).
2. So the triple product is \( 1\cdot[\vec a\ \vec b\ \vec c] \). For mutually orthogonal unit vectors \( [\vec a\ \vec b\ \vec c]=\pm1 \) (+1 if right-handed, −1 if left-handed).
3. So the value is 1 or −1.

Teacher's Note:
1. Triple product = determinant × [a b c].
2. Orthogonal unit vectors have [a b c] = ±1.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A non-zero vector perpendicular to a, b and c exists only if they are coplanar, so \( [\vec a\ \vec b\ \vec c]=0 \) (Statement-II is true).
2. In fact \( [\vec a-\vec b\quad\vec b-\vec c\quad\vec c-\vec a]=0 \) for any three vectors (the three vectors add to zero), so Statement-I is true by itself, and not because of Statement-II.
3. Both are true, but Statement-II is not the explanation.

Teacher's Note:
1. The vectors a − b, b − c, c − a sum to zero.
2. A non-zero r perpendicular to three vectors forces them to be coplanar.