JEE Mathematics Vector Algebra MCQs Set 16

Answer: (d) \( \tfrac32 \)

Explanation:
1. \( \vec a\cdot\vec b=10-3+2=9 \) and \( |\vec a|=3 \).
2. Projection of \( \tfrac12\vec b \) on \( \vec a \) \( =\dfrac{\tfrac12\,\vec a\cdot\vec b}{|\vec a|}=\dfrac{9}{2\cdot3}=\tfrac32 \).

Teacher's Note:
1. Projection of \( \vec b \) on \( \vec a \) is \( \dfrac{\vec a\cdot\vec b}{|\vec a|} \).
2. A scalar factor on \( \vec b \) multiplies the projection by the same factor.

Answer: (a) \( \dfrac{9(3\hat j+4\hat k)}{25} \)

Explanation:
1. \( \vec a\cdot\vec b=9 \) and \( |\vec b|^2=25 \).
2. The vector component of \( \vec a \) along \( \vec b \) is \( \dfrac{(\vec a\cdot\vec b)\vec b}{|\vec b|^2}=\dfrac{9(3\hat j+4\hat k)}{25} \).

Teacher's Note:
1. The vector component along \( \vec b \) is \( \dfrac{(\vec a\cdot\vec b)\vec b}{|\vec b|^2} \).
2. Divide by the square of the magnitude, not the magnitude itself.

Answer: (a) \( 4 \)

Explanation:
1. Projection of \( \vec a \) on \( \vec b \) \( =|\vec a|\cos\dfrac{5\pi}6=-\dfrac{\sqrt3}2|\vec a| \).
2. \( -\dfrac{\sqrt3}{2}|\vec a|=-\dfrac{6}{\sqrt3} \) gives \( |\vec a|=\dfrac{12}{3}=4 \).

Teacher's Note:
1. The projection equals the magnitude times the cosine of the angle.
2. \( \cos150^\circ \) is negative, so the signs agree.

Answer: (c) \( b\hat j \)

Explanation:
1. The component along the Y-axis is the part along \( \hat j \), namely \( \left((a\hat i+b\hat j+c\hat k)\cdot\hat j\right)\hat j=b\hat j \).

Teacher's Note:
1. The coefficient of \( \hat j \) gives the Y-component.
2. As a vector component it is \( b\hat j \); as a scalar it is \( b \).

Answer: (c) \( \sqrt3(\hat i+\hat j+\hat k) \)

Explanation:
1. Let the vector be \( t(\hat i+\hat j+\hat k) \). Its length is \( |t|\sqrt3=3 \), so \( t=\pm\sqrt3 \).
2. The vector \( \sqrt3(\hat i+\hat j+\hat k) \) is offered.

Teacher's Note:
1. Equal components means a multiple of \( \hat i+\hat j+\hat k \).
2. Use the length to find the multiplier.

Answer: (a) \( \dfrac{19}9 \)

Explanation:
1. \( (1,-2,1)\cdot(4,-4,7)=4+8+7=19 \) and \( |(4,-4,7)|=\sqrt{16+16+49}=9 \).
2. Length of the projection \( =\dfrac{19}{9} \).

Teacher's Note:
1. The length of a projection is a scalar, so the answer cannot be a vector.
2. Divide the dot product by the magnitude of the vector projected on.

Answer: (c) \( \dfrac{22}{\sqrt{133}} \)

Explanation:
1. \( 2\vec a-\vec b=(-1,3,0) \) and \( \vec a+\vec b=(4,-6,9) \) with \( |\vec a+\vec b|=\sqrt{16+36+81}=\sqrt{133} \).
2. Dot product \( =-4-18+0=-22 \), so the magnitude of the projection is \( \dfrac{22}{\sqrt{133}} \).

Teacher's Note:
1. Write the relevant formula first, then substitute the components carefully.
2. Check the arithmetic once more by an independent route.

Answer: (c) \( \tfrac13(\hat i-\hat j+\hat k) \)

Explanation:
1. \( \vec a\cdot\vec b=2+6+2=10 \) and \( |\vec b|^2=6 \), so the component along \( \vec b \) is \( \dfrac{10}{6}(1,2,1) \).
2. Perpendicular component \( =(2,3,2)-\dfrac53(1,2,1)=\left(\dfrac13,-\dfrac13,\dfrac13\right) \).

Teacher's Note:
1. Subtract the component along \( \vec b \) from the vector.
2. Check that the result is perpendicular to \( \vec b \).

Answer: (c) orthocentre

Explanation:
1. From the first equality, \( \vec b\cdot\vec c-\vec c\cdot\vec a=\vec b\cdot\vec r-\vec a\cdot\vec r \), i.e. \( \vec c\cdot(\vec b-\vec a)=\vec r\cdot(\vec b-\vec a) \), so \( (\vec r-\vec c)\cdot(\vec b-\vec a)=0 \).
2. So CP is perpendicular to AB; the other equality gives AP perpendicular to BC.
3. Two altitudes meet at P, so P is the orthocentre.

Teacher's Note:
1. A point whose joining lines to two vertices are perpendicular to the opposite sides is the orthocentre.
2. Rearrange each equality into a dot product equal to zero.

Answer: (b) \( 1 \)

Explanation:
1. \( |\vec a+\vec b+\vec c|^2=0=|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a) \).
2. Using the given relation, \( |\vec a|^2+|\vec b|^2+|\vec c|^2-3=0 \), so the sum of squares of lengths is 3.
3. With two lengths equal to 1, the third has length 1.

Teacher's Note:
1. Square the vector relation \( \vec a+\vec b+\vec c=\vec0 \).
2. Substitute the given dot-product relation.

Answer: (a) \( 1 \)

Explanation:
1. The sum is \( (2+x,\,6,\,-2) \). The unit vector along it dotted with \( (1,1,1) \) is \( \dfrac{x+6}{\sqrt{(x+2)^2+40}}=1 \).
2. Squaring: \( (x+6)^2=(x+2)^2+40 \), i.e. \( 8x+32=40 \), so \( x=1 \).

Teacher's Note:
1. The dot product with a unit vector equals the projection.
2. Square both sides to remove the root and solve the linear equation.

Answer: (b) \( 8 \)

Explanation:
1. \( x+y+2=3 \) and \( 2x-y+3=5 \), so \( x+y=1 \) and \( 2x-y=2 \). Adding: \( x=1 \), then \( y=0 \).
2. \( \vec r\cdot\hat k=x+2y+7=1+0+7=8 \).

Teacher's Note:
1. Dot products with \( \hat i,\hat j,\hat k \) pick out the components.
2. Solve the two equations for x and y first.

Answer: (d) \( 4 \)

Explanation:
1. Let \( \overrightarrow{AB}=\vec p \) and \( \overrightarrow{AD}=\vec q \). \( \overrightarrow{AC}=\vec p+\vec q \), \( \overrightarrow{BD}=\vec q-\vec p \).
2. \( AC^2-BD^2=(|\vec p|^2+2\vec p\cdot\vec q+|\vec q|^2)-(|\vec p|^2-2\vec p\cdot\vec q+|\vec q|^2)=4\,\vec p\cdot\vec q \). So \( K=4 \).

Teacher's Note:
1. Write both diagonals using two adjacent sides.
2. Subtract the two squared lengths; the squares cancel.

Answer: (c) \( -1,\ 2 \)

Explanation:
1. Adjacent sides of a rectangle are perpendicular: \( 2a^2-2a-4=0 \).
2. \( a^2-a-2=0 \), so \( (a-2)(a+1)=0 \) and \( a=2 \) or \( a=-1 \).

Teacher's Note:
1. Adjacent sides of a rectangle are perpendicular.
2. Set the dot product equal to zero and factorise.

Answer: (d) \( 3 \)

Explanation:
1. \( |\vec a+\vec b|^2+|\vec a-\vec b|^2=2(|\vec a|^2+|\vec b|^2) \).
2. \( 1+49=2(16+|\vec b|^2) \), so \( |\vec b|^2=9 \) and \( |\vec b|=3 \).

Teacher's Note:
1. Add the two expansions to cancel the dot product.
2. This is the parallelogram law.

Answer: (d) \( (4,-2,4) \)

Explanation:
1. Let \( \vec b=\lambda\vec a \). Then \( \vec a\cdot\vec b=\lambda|\vec a|^2=9\lambda=18 \), so \( \lambda=2 \).
2. \( \vec b=(4,-2,4) \).

Teacher's Note:
1. A collinear vector is a scalar multiple.
2. Use the dot product to find the multiplier.

Answer: (a) \( 2\vec a \)

Explanation:
1. For an orthonormal triple, \( \vec r=(\vec r\cdot\vec a)\vec a+(\vec r\cdot\vec b)\vec b+(\vec r\cdot\vec c)\vec c \).
2. So the expression equals \( (\vec r\cdot\vec a)\vec a=2\vec a \).

Teacher's Note:
1. Any vector can be expanded along an orthonormal triple using dot products.
2. Remove the two other components to be left with the first.

Answer: (c) \( \dfrac{2\pi}3<\theta\le\pi \)

Explanation:
1. \( |\vec a+\vec b|^2=2+2\cos\theta<1 \), so \( \cos\theta<-\dfrac12 \).
2. This holds for \( \dfrac{2\pi}3<\theta\le\pi \).

Teacher's Note:
1. Square the magnitude and use the dot product.
2. \( \cos\theta<-\tfrac12 \) means an angle beyond \( 120^\circ \).

Answer: (a) \( \dfrac\pi3 \)

Explanation:
1. \( |\vec a-\vec b|^2=4+1-2\vec a\cdot\vec b=3 \), so \( \vec a\cdot\vec b=1 \).
2. \( \cos\theta=\dfrac{1}{2\cdot1}=\dfrac12 \), so \( \theta=\dfrac\pi3 \).

Teacher's Note:
1. Expand \( |\vec a-\vec b|^2 \) to find the dot product.
2. Then use \( \cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|} \).

Answer: (a) \( \dfrac\pi2 \)

Explanation:
1. The dot product is \( 8(1+0-1)=0 \).
2. The vectors are perpendicular, so the angle is \( \dfrac\pi2 \).

Teacher's Note:
1. A zero dot product means the vectors are perpendicular.
2. The scalar factor 8 does not change the angle.

Answer: (d) \( \dfrac\pi2 \)

Explanation:
1. Squaring: \( |\alpha|^2+2\alpha\cdot\beta+|\beta|^2=|\alpha|^2-2\alpha\cdot\beta+|\beta|^2 \), so \( \vec\alpha\cdot\vec\beta=0 \).
2. The vectors are perpendicular, and so are \( -\vec\alpha \) and \( \vec\beta \): the angle is \( \dfrac\pi2 \).

Teacher's Note:
1. Equal magnitudes of sum and difference mean perpendicular vectors.
2. Reversing one vector keeps them perpendicular.

Answer: (a) \( \cos^{-1}\dfrac23,\cos^{-1}\dfrac23,\cos^{-1}\dfrac13 \)

Explanation:
1. Direction ratios: \( (2,-2,-1) \), magnitude \( 3 \). Direction cosines (in absolute value): \( \dfrac23,\dfrac23,\dfrac13 \).
2. The acute angles are \( \cos^{-1}\dfrac23,\ \cos^{-1}\dfrac23,\ \cos^{-1}\dfrac13 \).

Teacher's Note:
1. The acute angle uses the absolute value of each direction cosine.
2. A cosine larger than 1 is impossible, which eliminates two options.

Answer: (d) \( (-4,7) \)

Explanation:
1. The dot product is \( x^2-3x-28 \), and it must be negative.
2. \( (x-7)(x+4)<0 \) gives \( -4<x<7 \).

Teacher's Note:
1. Obtuse means the dot product is negative (and the vectors are not opposite).
2. Factorise the quadratic and read the interval between the roots.

Answer: (b) \( a^2=b^2+c^2 \)

Explanation:
1. \( |\vec a|^2=|\vec b|^2+|\vec c|^2+2\vec b\cdot\vec c \) and \( \vec b\cdot\vec c=0 \).
2. So \( a^2=b^2+c^2 \).

Teacher's Note:
1. This is the Pythagoras theorem for perpendicular vectors.
2. The cross term vanishes when the dot product is zero.

Answer: (c) \( \dfrac{35}{41} \)

Explanation:
1. \( \overrightarrow{AB}=(-1,-2,-6) \) and \( \overrightarrow{AC}=(1,-3,-5) \). \( \overrightarrow{AB}\cdot\overrightarrow{AC}=-1+6+30=35 \).
2. \( |AB|^2=41 \), \( |AC|^2=35 \). \( \cos^2A=\dfrac{35^2}{41\cdot35}=\dfrac{35}{41} \).

Teacher's Note:
1. Angle A is between the vectors AB and AC.
2. Squaring the cosine avoids the root.

Answer: (b) \( \dfrac\pi2 \)

Explanation:
1. \( \vec a+\vec b=(4,1,-1) \) and \( \vec a-\vec b=(-2,3,-5) \).
2. Dot product \( =-8+3+5=0 \), so the angle is \( \dfrac\pi2 \).

Teacher's Note:
1. Compute the sum and the difference first.
2. A zero dot product means a right angle.