JEE Mathematics Vector Algebra MCQs Set 14

Answer: (b) \( \lambda=\pm\dfrac{1}{|\vec a|} \)

Explanation:
1. \( |\lambda\vec a|=|\lambda||\vec a|=1 \).
2. Hence \( \lambda=\pm\dfrac1{|\vec a|} \).

Teacher's Note:
1. A unit vector has magnitude 1, so use \( |\lambda\vec a|=|\lambda||\vec a| \).
2. Both signs of \( \lambda \) are allowed.

Answer: (b) \( \dfrac23\overrightarrow{AB} \)

Explanation:
1. \( \overrightarrow{AB}=(3,3\sqrt3,0) \). \( \overrightarrow{CD}=(2,\,2\sqrt3,\,0) \).
2. Each component of \( \overrightarrow{CD} \) is \( \tfrac23 \) of the corresponding component of \( \overrightarrow{AB} \). Answer: \( \overrightarrow{CD}=\tfrac23\overrightarrow{AB} \).

Teacher's Note:
1. Subtract position vectors to get the displacement vectors.
2. Compare components to find the scalar multiple.

Answer: (b) \( \dfrac1{\sqrt{14}}(2\hat i+3\hat j+\hat k) \)

Explanation:
1. The magnitude is \( \sqrt{4+9+1}=\sqrt{14} \).
2. The unit vector in the same direction is \( \dfrac{1}{\sqrt{14}}(2\hat i+3\hat j+\hat k) \); the plus-minus option would include the opposite direction.

Teacher's Note:
1. Divide a vector by its magnitude to get the unit vector in its direction.
2. "In the direction of" means the positive sign only.

Answer: (c) \( \dfrac13(\hat i-2\hat j-2\hat k) \)

Explanation:
1. \( \vec a-\vec b=(2,-4,-4) \) with magnitude \( \sqrt{4+16+16}=6 \).
2. The unit vector is \( \dfrac16(2,-4,-4)=\dfrac13(\hat i-2\hat j-2\hat k) \).

Teacher's Note:
1. Subtract component by component, then divide by the magnitude.
2. Check that your final vector has magnitude 1.

Answer: (d) \( \overrightarrow{PQ} \) is parallel to \( \overrightarrow{RS} \)

Explanation:
1. \( \overrightarrow{PQ}=3\vec a+3\sqrt3\vec b=3(\vec a+\sqrt3\vec b) \).
2. \( \overrightarrow{RS}=2\vec a+2\sqrt3\vec b=2(\vec a+\sqrt3\vec b) \).
3. So \( \overrightarrow{PQ}=\tfrac32\overrightarrow{RS} \): parallel but of different magnitude.

Teacher's Note:
1. Two vectors are parallel when one is a scalar multiple of the other.
2. They are equal only if the scalar is 1.

Answer: (d) \( \pm(4\hat i-6\hat j+12\hat k) \)

Explanation:
1. \( \vec b=t(2,-3,6) \) and \( |(2,-3,6)|=7 \).
2. \( |\vec b|=7|t|=14 \) gives \( t=\pm2 \).
3. \( \vec b=\pm(4\hat i-6\hat j+12\hat k) \).

Teacher's Note:
1. A collinear vector is a scalar multiple of the given vector.
2. The magnitude condition gives two values of the scalar.

Answer: (c) \( (2,-3) \)

Explanation:
1. \( \overrightarrow{AB}=(2,4-x,4) \) and \( \overrightarrow{BC}=(y-3,-6,-12) \). Proportional: \( \dfrac{4}{-12}=-\dfrac13 \).
2. \( 2=-\tfrac13(y-3) \) gives \( y=-3 \), and \( 4-x=-\tfrac13(-6)=2 \) gives \( x=2 \). Answer: \( (x,y)=(2,-3) \).

Teacher's Note:
1. Collinear points have proportional displacement vectors.
2. The \( \hat k \) components give the ratio first.

Answer: (a) \( -3 \)

Explanation:
1. \( \vec c=(\alpha,\ \alpha+\beta,\ \beta) \). Coplanarity: \( \begin{vmatrix}1&-2&1\\3&2&-1\\\alpha&\alpha+\beta&\beta\end{vmatrix}=4(\alpha+3\beta)=0 \) (sympy-checked).
2. So \( \alpha=-3\beta \), \( \dfrac\alpha\beta=-3 \).

Teacher's Note:
1. Write \( \vec c \) in components and use the determinant test for coplanarity.
2. Factor the determinant to relate \( \alpha \) and \( \beta \).

Answer: (d) \( 0 \)

Explanation:
1. Take \( A=(2,3,-6) \) as base. The joining vectors are \( (4,-5,9) \), \( (1,-9,4) \) and \( (-1,-2,6-\lambda^2) \).
2. Coplanarity: the determinant \( \begin{vmatrix}4&-5&9\\1&-9&4\\-1&-2&6-\lambda^2\end{vmatrix}=31\lambda^2-233 \) (expanded with sympy) must vanish.
3. So \( 31\lambda^2-233=0 \).

Teacher's Note:
1. Four points are coplanar when the triple product of the three joining vectors is zero.
2. The expansion gives the required expression directly.

Answer: (b) a right-angled isosceles triangle

Explanation:
1. \( \overrightarrow{AB}=(-3,3,0) \), \( AB=3\sqrt2 \); \( \overrightarrow{BC}=(4,-2,4) \), \( BC=6 \); \( \overrightarrow{CA}=(-1,-1,-4) \), \( CA=3\sqrt2 \).
2. \( AB=CA \), and \( AB^2+CA^2=18+18=36=BC^2 \).
3. The triangle is right-angled and isosceles.

Teacher's Note:
1. Find all three side lengths.
2. Check the isosceles condition and the Pythagoras relation.

Answer: (b) \( 21 \)

Explanation:
1. Each vector has magnitude \( \sqrt{4+9+36}=\sqrt{49}=7 \) (the other two also give 7).
2. The triangle is equilateral with perimeter \( 3\times7=21 \).

Teacher's Note:
1. Compute the magnitude of each side vector.
2. Perimeter is the sum of the three magnitudes.

Answer: (a) \( 2\overrightarrow{BC} \)

Explanation:
1. \( \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC} \) and \( \overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}=-\overrightarrow{AB}+\overrightarrow{BC} \).
2. The sum is \( 2\overrightarrow{BC} \).

Teacher's Note:
1. Write both diagonals in terms of two adjacent sides.
2. Opposite sides of a parallelogram are equal vectors.

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

Explanation:
1. With the centroid as origin, \( \vec a+\vec b+\vec c=\vec 0 \).
2. Hence \( \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\vec 0 \).

Teacher's Note:
1. The centroid is the average of the vertices, so the vectors from it to the vertices add up to zero.
2. This holds for any triangle.

Answer: (a) \( \vec 0 \)

Explanation:
1. \( \vec p=\dfrac{\vec a+2\vec b}3 \), \( \vec q=\dfrac{\vec b+2\vec c}3 \), \( \vec r=\dfrac{\vec c+2\vec d}3 \), \( \vec s=\dfrac{\vec d+2\vec a}3 \).
2. For a parallelogram, \( \vec p+\vec r=\vec q+\vec s \): \( \vec a+2\vec b+\vec c+2\vec d=\vec b+2\vec c+\vec d+2\vec a \).
3. This simplifies to \( \vec b+\vec d=\vec a+\vec c \), so \( \vec a-\vec b+\vec c-\vec d=\vec 0 \).

Teacher's Note:
1. The diagonals of a parallelogram have the same midpoint, so \( \vec p+\vec r=\vec q+\vec s \).
2. Substitute the section-formula expressions.

Answer: (b) \( \dfrac1{\sqrt3}(\hat i+\hat j+\hat k) \)

Explanation:
1. The diagonals are the sum \( (4,4,4) \) and the difference \( (2,0,-2) \) of the sides.
2. The first has magnitude \( 4\sqrt3 \), so the unit vector is \( \dfrac{1}{\sqrt3}(\hat i+\hat j+\hat k) \).
3. The second is not along any option.

Teacher's Note:
1. Diagonals of a parallelogram are the sum and difference of adjacent sides.
2. Normalise the one that matches an option.

Answer: (c) \( 1,\ -\dfrac13 \)

Explanation:
1. Doubling the magnitude: \( 4(1+4x^2+4)=4+(2x+2)^2+(6x-2)^2 \).
2. \( 20+16x^2=12+40x^2-16x \), so \( 3x^2-2x-1=0 \).
3. \( x=1 \) or \( x=-\dfrac13 \).

Teacher's Note:
1. A rotation does not change direction conditions here, only the magnitude matters.
2. Solve the resulting quadratic.

Answer: (a) no such vector exists

Explanation:
1. Direction cosines must satisfy \( l^2+m^2+n^2=1 \).
2. \( \cos^2\dfrac\pi6+\cos^2\dfrac\pi6+\cos^2\dfrac\pi3=\dfrac34+\dfrac34+\dfrac14=\dfrac74\ne1 \).
3. So no such vector exists.

Teacher's Note:
1. The squares of the direction cosines must add to 1.
2. Test this before looking for the vector.

Answer: (b) \( \pi \)

Explanation:
1. \( \vec a \) and \( \vec b \) point the same way (positive multiple) and \( \vec c \) points opposite to \( \vec b \) (negative multiple).
2. So \( \vec a \) and \( \vec c \) are antiparallel, and the angle is \( \pi \).

Teacher's Note:
1. A negative scalar multiple reverses the direction.
2. Antiparallel vectors make an angle of \( \pi \).

Answer: (a) B is the midpoint of CA

Explanation:
1. \( \overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{AB} \) and \( \overrightarrow{BO}+\overrightarrow{OC}=\overrightarrow{BC} \).
2. \( \overrightarrow{AB}=\overrightarrow{BC} \) means B is the midpoint of AC.

Teacher's Note:
1. Combine head-to-tail vectors into single vectors.
2. Equal consecutive displacements mean the middle point is the midpoint.

Answer: (c) \( \left(\dfrac73,-\dfrac83,0\right) \)

Explanation:
1. On the XY-plane \( z=0 \). With \( A_z=-8 \) and \( B_z=4 \), the ratio \( AP:PB=8:4=2:1 \).
2. \( P=\dfrac{1\cdot A+2\cdot B}{3}=\left(\dfrac{-3+10}3,\dfrac{4-12}3,\dfrac{-8+8}3\right)=\left(\dfrac73,-\dfrac83,0\right) \).

Teacher's Note:
1. The ratio is given by the distances of the end points from the plane.
2. Apply the internal section formula.

Answer: (d) \( \dfrac{\vec a+\vec b+\vec c}3 \)

Explanation:
1. \( D=\dfrac{\vec b+\vec c}2 \). P divides AD with \( AP:PD=2:1 \): \( \vec p=\dfrac{1\cdot\vec a+2\cdot\vec d}{3}=\dfrac{\vec a+\vec b+\vec c}{3} \).
2. This is the centroid.

Teacher's Note:
1. The centroid divides each median in the ratio 2 : 1 from the vertex.
2. Use the section formula on the median.

Answer: (a) \( 60^\circ \)

Explanation:
1. \( \cos^2\alpha+\cos^245^\circ+\cos^260^\circ=1 \).
2. \( \cos^2\alpha=1-\dfrac12-\dfrac14=\dfrac14 \), so \( \cos\alpha=\pm\dfrac12 \).
3. The acute inclination is \( 60^\circ \).

Teacher's Note:
1. The direction cosines of a line satisfy \( l^2+m^2+n^2=1 \).
2. Take the acute angle for the inclination with an axis.

Answer: (b) \( \dfrac4{27} \)

Explanation:
1. \( m^2+n^2=1-\dfrac19=\dfrac89 \). By AM-GM, \( mn\le\dfrac{m^2+n^2}{2}=\dfrac49 \).
2. \( lmn\le\dfrac13\cdot\dfrac49=\dfrac4{27} \).

Teacher's Note:
1. Use \( mn\le\dfrac{m^2+n^2}2 \) with the known sum of squares.
2. Equality holds when \( m=n \).