JEE Mathematics Vector Algebra MCQs Set 17

Answer: (b) \( 7 \)

Explanation:
1. \( |\vec a-\vec b|^2=9+25-2\cdot3\cdot5\cos120^\circ=34+15=49 \).
2. So \( |\vec a-\vec b|=7 \).

Teacher's Note:
1. Use the cosine rule form \( |\vec a-\vec b|^2=a^2+b^2-2ab\cos\theta \).
2. \( \cos120^\circ=-\tfrac12 \) turns the minus into a plus.

Answer: (d) \( \cos^{-1}\dfrac4{21} \)

Explanation:
1. \( \vec a\cdot\vec b=12-6-2=4 \), \( |\vec a|=3 \), \( |\vec b|=7 \).
2. \( \cos\theta=\dfrac{4}{21} \).

Teacher's Note:
1. Use \( \cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|} \).
2. Compute both magnitudes before dividing.

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

Explanation:
1. \( (\vec e_1+2\vec e_2)\cdot(5\vec e_1-4\vec e_2)=5-4c+10c-8=0 \), where \( c=\vec e_1\cdot\vec e_2 \).
2. \( 6c=3 \), so \( c=\tfrac12 \) and the angle is \( 60^\circ \).

Teacher's Note:
1. Expand the dot product of the two perpendicular vectors.
2. Use \( \vec e_i\cdot\vec e_i=1 \) for unit vectors.

Answer: (d) \( \dfrac\pi3 \)

Explanation:
1. \( |\vec a+\vec b|=\sqrt3|\vec c|=\sqrt3 \), so \( 2+2\vec a\cdot\vec b=3 \) and \( \vec a\cdot\vec b=\tfrac12 \).
2. The angle is \( \dfrac\pi3 \).

Teacher's Note:
1. Move the term with \( \vec c \) to one side before squaring.
2. For unit vectors the dot product is the cosine of the angle.

Answer: (b) \( \cos^{-1}\dfrac{12}{77} \)

Explanation:
1. \( \vec a\cdot\vec b=12-18+18=12 \), \( |\vec a|=7 \), \( |\vec b|=11 \).
2. \( \cos\theta=\dfrac{12}{77} \).

Teacher's Note:
1. Compute the dot product and both magnitudes.
2. \( 6^2+2^2+3^2=49 \) and \( 2^2+9^2+6^2=121 \).

Answer: (a) \( \dfrac25 \)

Explanation:
1. Let \( |\vec a|=p \), so \( |\vec b|=2p \) and \( \vec a\cdot\vec b=p\cdot2p\cos120^\circ=-p^2 \).
2. \( (\vec a+x\vec b)\cdot(\vec a-\vec b)=p^2-(-p^2)+x(-p^2)-x(4p^2)=2p^2-5xp^2=0 \), so \( x=\dfrac25 \).

Teacher's Note:
1. Express every dot product in terms of \( |\vec a|^2 \).
2. Set the dot product of the perpendicular vectors to zero.

Answer: (b) \( 45^\circ \)

Explanation:
1. D is the midpoint of AC: \( D=(1,-1,2) \), so \( \overrightarrow{BD}=(-1,0,1) \). \( \overrightarrow{AC}=(-2,4,4) \).
2. \( \cos\theta=\dfrac{2+0+4}{\sqrt2\cdot6}=\dfrac1{\sqrt2} \), so \( \theta=45^\circ \).

Teacher's Note:
1. Locate the midpoint D before forming BD.
2. The angle between the lines is the angle between their direction vectors.

Answer: (d) \( 4x+7y-4z+6=0 \)

Explanation:
1. The plane is \( 4x+7y-4z=d \). Substituting \( (3,-2,1) \): \( 12-14-4=-6 \).
2. So \( 4x+7y-4z+6=0 \).

Teacher's Note:
1. The normal vector gives the coefficients of x, y, z.
2. Substitute the given point to find the constant.

Answer: (d) \( \dfrac{3}{5\sqrt2} \)

Explanation:
1. Distance \( =\dfrac{|3|}{\sqrt{9+16+25}}=\dfrac3{\sqrt{50}} \).
2. \( \sqrt{50}=5\sqrt2 \), so the distance is \( \dfrac3{5\sqrt2} \).

Teacher's Note:
1. Distance from the origin is the constant divided by the length of the normal.
2. Simplify the root of 50.

Answer: (a) \( \dfrac5{\sqrt2} \)

Explanation:
1. Distance \( =\dfrac{25}{\sqrt{9+16+25}}=\dfrac{25}{5\sqrt2}=\dfrac5{\sqrt2} \).

Teacher's Note:
1. Divide the constant by the magnitude of the normal vector.
2. Check that \( \sqrt{50}=5\sqrt2 \).

Answer: (c) \( \cos^{-1}\dfrac13 \)

Explanation:
1. Normals \( (1,-1,1) \) and \( (1,1,1) \): dot product \( 1-1+1=1 \), each magnitude \( \sqrt3 \).
2. \( \cos\theta=\dfrac13 \).

Teacher's Note:
1. Take the normals of both planes from the coefficients.
2. Do not confuse the angle between planes with the angle between a line and a plane.

Answer: (b) \( 2 \)

Explanation:
1. The displacement is \( \overrightarrow{AB}=(0,2,0) \).
2. Work \( =\vec F\cdot\overrightarrow{AB}=0+2+0=2 \) units.

Teacher's Note:
1. Work is the dot product of force and displacement.
2. Find the displacement as final minus initial position.

Answer: (d) \( 3 \)

Explanation:
1. The displacement is \( (1,1,1) \).
2. Work \( =a+1+1=5 \), so \( a=3 \).

Teacher's Note:
1. Work equals force dot displacement.
2. Solve the resulting linear equation.

Answer: (b) \( \dfrac{\hat i+4\hat j+\hat k}3 \)

Explanation:
1. \( (1,1,1)\cdot(2,-1,2)=3 \) and \( |(2,-1,2)|^2=9 \), so the component along it is \( \dfrac39(2,-1,2) \).
2. Perpendicular component \( =(1,1,1)-\left(\dfrac23,-\dfrac13,\dfrac23\right)=\left(\dfrac13,\dfrac43,\dfrac13\right)=\dfrac{\hat i+4\hat j+\hat k}3 \).

Teacher's Note:
1. Subtract the component along the given vector.
2. The result must have zero dot product with the given vector.

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

Explanation:
1. The projection of \( \vec b \) on \( \vec a \) and the component of \( \vec b \) perpendicular to \( \vec a \) add up to \( \vec b \) itself.
2. The unit vector along \( \vec b \) is \( \dfrac{\hat i+2\hat j+3\hat k}{\sqrt{14}} \).

Teacher's Note:
1. A vector splits into a part along a given direction and a part perpendicular to it.
2. Their sum is the original vector.

Answer: (c) \( 0 \)

Explanation:
1. The expression is \( |\vec a+\vec b|^2-|\vec c|^2 \).
2. Since \( \vec a+\vec b=-\vec c \), this is \( |\vec c|^2-|\vec c|^2=0 \), a scalar zero (not the null vector).

Teacher's Note:
1. Recognise \( |\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b \) as \( |\vec a+\vec b|^2 \).
2. A dot-product expression is a scalar, never a vector.

Answer: (b) \( \sqrt{593} \)

Explanation:
1. \( \vec a\cdot\vec b=2\sqrt2\cdot3\cdot\dfrac1{\sqrt2}=6 \). The diagonals are the sum \( 6\vec a-\vec b \) and the difference \( 4\vec a+5\vec b \).
2. \( |6\vec a-\vec b|^2=36\cdot8+9-12\cdot6=225 \) and \( |4\vec a+5\vec b|^2=16\cdot8+25\cdot9+40\cdot6=593 \).
3. The longer diagonal has length \( \sqrt{593} \).

Teacher's Note:
1. The diagonals are the sum and the difference of the adjacent sides.
2. Compare the two squared lengths to decide which is longer.

Answer: (a) \( -334 \)

Explanation:
1. \( \vec a\cdot\vec b=1 \), \( \vec a\cdot\vec c=\tfrac32 \), \( \vec b\cdot\vec c=3 \), and \( |\vec a|^2=1,\ |\vec b|^2=4,\ |\vec c|^2=9 \).
2. Expanding: \( 8-12+30+12-72+90-30+90-450=-334 \).

Teacher's Note:
1. Expand the product term by term using the pairwise dot products.
2. Keep a table of the six dot products to avoid mistakes.

Answer: (c) \( \dfrac{\vec a+\vec b+\vec c}{\sqrt3} \)

Explanation:
1. For perpendicular vectors of equal length, \( (\vec a+\vec b+\vec c)\cdot\vec a=(\vec a+\vec b+\vec c)\cdot\vec b=(\vec a+\vec b+\vec c)\cdot\vec c \), so \( \vec a+\vec b+\vec c \) makes equal angles with all three.
2. The options with minus signs make unequal angles.

Teacher's Note:
1. Check the dot product of the candidate with each of the three vectors.
2. A sign change in one term breaks the equality of the angles.

Answer: (d) angle BCA is a right angle

Explanation:
1. At C: \( \overrightarrow{CB}=(8,1,4) \) and \( \overrightarrow{CA}=(-4,4,7) \). Dot product \( =-32+4+28=0 \).
2. So angle BCA is a right angle. At D: \( (3,6,6)\cdot(-1,10,13)=135\ne0 \).

Teacher's Note:
1. Test each angle with the dot product of the two vectors from its vertex.
2. The vertex is the middle letter of the angle name.

Answer: (a) \( -\dfrac12BC^2 \)

Explanation:
1. Let the side be \( s \). The interior angle at A is \( 120^\circ \), so \( \overrightarrow{AB}\cdot\overrightarrow{AF}=s^2\cos120^\circ=-\dfrac{s^2}2 \).
2. \( BC=s \), so the value is \( -\dfrac12BC^2 \).

Teacher's Note:
1. Use the interior angle of a regular hexagon.
2. \( \cos120^\circ=-\tfrac12 \).

Answer: (d) \( \hat i+\hat k \)

Explanation:
1. \( |\vec a|=|\vec b|=\sqrt2 \) and \( \vec a\cdot\vec b=1 \), so each pair has dot product 1 and \( |\vec c|^2=2 \). With \( \vec c=(x,y,z) \): \( x+y=1,\ y+z=1,\ x^2+y^2+z^2=2 \).
2. This gives \( x=z=1-y \) and \( 3y^2-4y=0 \), so \( y=0 \) (the other root \( y=\tfrac43 \) gives a vector that is not offered). \( \vec c=\hat i+\hat k \).

Teacher's Note:
1. Equal lengths and equal angles mean all pairwise dot products are equal.
2. Solve the resulting equations for the components.

Answer: (d) \( -\tfrac32\hat i+\tfrac52\hat j+3\hat k \)

Explanation:
1. Let \( \vec c=\lambda\vec a+\mu\vec b \). \( \vec a\cdot\vec b=-1 \), \( |\vec b|^2=5 \), \( |\vec a|^2=3 \). \( \vec c\cdot\vec b=0 \) gives \( -\lambda+5\mu=0 \), so \( \mu=\dfrac\lambda5 \).
2. \( \vec a\cdot\vec c=3\lambda-\mu=\dfrac{14\lambda}5=7 \), so \( \lambda=\dfrac52 \), \( \mu=\dfrac12 \).
3. \( \vec c=\tfrac52(-1,1,1)+\tfrac12(2,0,1)=\left(-\tfrac32,\tfrac52,3\right) \).

Teacher's Note:
1. Write the coplanar vector as a combination of the two given vectors.
2. Use the two dot-product conditions to find the coefficients.

Answer: (a) are mutually perpendicular

Explanation:
1. The first two vectors are \( \vec a=(1,1,1) \) and \( \vec b=(1,-1,0) \), and the third is \( (1,1,-2) \).
2. Dot products: \( 1-1+0=0 \), \( 1+1-2=0 \), \( 1-1+0=0 \). They are mutually perpendicular.
3. The volume is \( \sqrt3\cdot\sqrt2\cdot\sqrt6=6 \), so it is not coplanar and not 2 or 3.

Teacher's Note:
1. Each bracket expression simply rebuilds the vector in components.
2. Test all three pairs by dot products.

Answer: (b) \( \dfrac{\vec a+\vec c}2 \)

Explanation:
1. \( (\vec a-\vec b)\cdot(\vec c-\vec b)=\overrightarrow{BA}\cdot\overrightarrow{BC}=0 \), so the angle at B is \( 90^\circ \).
2. The circumcentre of a right-angled triangle is the midpoint of the hypotenuse AC: \( \dfrac{\vec a+\vec c}2 \).

Teacher's Note:
1. Find which vertex carries the right angle.
2. The hypotenuse is a diameter of the circumcircle.