JEE Mathematics Vector Algebra MCQs Set 12

Answer: (b) (i) and (iii) only

Explanation:
1. \( \vec a\perp\vec b \), \( \vec b\perp\vec c \), so \( |\vec c|=|\vec a||\vec b| \) and \( |\vec a|=|\vec b||\vec c| \).
2. Substituting: \( |\vec c|=|\vec b|^2|\vec c| \), so \( |\vec b|=1 \), and then \( |\vec a|=|\vec c| \).
3. (ii) and (iv) need not hold, since \( |\vec a| \) can be any positive value. Answer: (i) and (iii).

Teacher's Note:
1. Perpendicular vectors give cross-product magnitudes equal to the product of lengths.
2. Combine the two magnitude equations.

Answer: (d) (i), (ii) and (iii) only

Explanation:
1. Since \( |\vec x|^2=2 \) and the angle is \( 60^\circ \), \( \vec x\cdot\vec y=\vec y\cdot\vec z=\vec z\cdot\vec x=1 \).
2. \( \vec a\parallel\vec x\times(\vec y\times\vec z)=\vec y(\vec x\cdot\vec z)-\vec z(\vec x\cdot\vec y)=\vec y-\vec z \), so \( \vec a=k(\vec y-\vec z) \) with \( \vec a\cdot\vec y=k(2-1)=k \): this is (ii). Likewise \( \vec b\parallel\vec y\times(\vec z\times\vec x)=\vec z-\vec x \) gives (i).
3. \( \vec a\cdot\vec b=kk'(\vec y-\vec z)\cdot(\vec z-\vec x)=kk'(1-1-2+1)=-kk' \), which is (iii); (iv) has the wrong sign. Answer: (i), (ii) and (iii).

Teacher's Note:
1. A vector perpendicular to \( \vec x \) and \( \vec y\times\vec z \) is along \( \vec x\times(\vec y\times\vec z) \).
2. Use \( \vec x\cdot\vec y=\vec x\cdot\vec z=1 \) for these vectors.

Answer: (b) \( 1 \) or \( -\dfrac13 \)

Explanation:
1. The magnitude does not change: \( 4p^2+1=(p+1)^2+1 \).
2. \( 3p^2-2p-1=0 \), so \( (3p+1)(p-1)=0 \).
3. \( p=1 \) or \( p=-\dfrac13 \).

Teacher's Note:
1. Rotation of axes preserves the length of a vector.
2. Equate the squares of the lengths before and after.

Answer: (c) (i) and (iii) only

Explanation:
1. With the angle \( \theta \) between the vectors, \( |\vec v|=\sin\theta \) and \( |\vec u|^2=1-2\cos^2\theta+\cos^2\theta=\sin^2\theta \), so \( |\vec u|=|\vec v| \).
2. \( \vec u\cdot\vec b=\cos\theta-\cos\theta=0 \), so (iii) holds.
3. \( \vec u\cdot\vec a=\sin^2\theta\ne0 \), so (ii) and (iv) fail. Answer: (i) and (iii).

Teacher's Note:
1. Note that \( \vec u \) is the component of \( \vec a \) perpendicular to \( \vec b \).
2. Its dot product with \( \vec b \) is therefore zero.

Answer: (a) (ii) and (iv) only

Explanation:
1. The plane is spanned by \( (1,1,1) \) and \( (-1,1,1) \), so its vectors are of the form \( (x,y,y) \).
2. Perpendicular to \( (1,-1,-1) \): \( x-y-y=0 \), so \( x=2y \).
3. (ii) \( (2,1,1) \) and (iv) \( (-4,-2,-2) \) fit; (i) and (iii) do not. Answer: (ii) and (iv).

Teacher's Note:
1. The vectors in this plane have equal \( \hat j \) and \( \hat k \) components.
2. Apply the perpendicularity condition to the general form.

Answer: (c) \( \dfrac\pi4 \)

Explanation:
1. Normal to \( P_1 \): \( (0,2,3)\times(0,4,-3)=(-18,0,0)\parallel\hat i \). Normal to \( P_2 \): \( (0,1,-1)\times(3,3,0)=(3,-3,-3)\parallel(1,-1,-1) \).
2. \( \vec A\parallel\hat i\times(1,-1,-1)=(0,1,-1) \).
3. \( \cos\phi=\dfrac{(0,1,-1)\cdot(2,1,-2)}{\sqrt2\cdot3}=\dfrac{3}{3\sqrt2}=\dfrac1{\sqrt2} \), so the angle is \( \dfrac\pi4 \).

Teacher's Note:
1. The line of intersection is along the cross product of the two normals.
2. Use the dot product formula for the angle.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. The coefficients \( 2+3-5=0 \), so \( \vec c=\dfrac{2\vec a+3\vec b}{5} \), a point dividing PQ in the ratio 3:2. Thus P, Q, R are collinear and A is true.
2. R is a true statement and is exactly the test of collinearity, so it explains A.

Teacher's Note:
1. A linear relation among position vectors with coefficients summing to zero shows collinearity.
2. Collinearity means one side vector is a scalar multiple of another.

Answer: (d) Both Assertion and Reason are false.

Explanation:
1. \( |\vec u+\vec v|=2\cos(\alpha/2) \), so the unit bisector is \( \dfrac{\vec u+\vec v}{2\cos(\alpha/2)} \). The sine in the assertion is wrong, so A is false.
2. In an isosceles triangle the midpoint D of BC lies on the bisector of angle A, so \( \overrightarrow{AD}=\tfrac12(\overrightarrow{AB}+\overrightarrow{AC}) \) is correct, and R is true.

Teacher's Note:
1. The diagonal of a rhombus of unit vectors bisects the angle between them.
2. Its length is \( 2\cos(\alpha/2) \).

Answer: (b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.

Explanation:
1. The incentre is \( \dfrac{a\,\vec A+b\,\vec B+c\,\vec C}{a+b+c} \), so \( a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC}=\vec 0 \) and A is true.
2. The centroid formula is also true, but it does not explain the incentre relation.

Teacher's Note:
1. The incentre weights the vertices by the opposite side lengths.
2. The centroid weights them equally, so it does not explain this result.

Answer: (a) inside triangle OAB

Explanation:
1. Here \( p=\dfrac12>0 \), \( q=\dfrac13>0 \) and \( p+q=\dfrac56<1 \).
2. These satisfy the conditions for a point inside the triangle.

Teacher's Note:
1. Check the signs of \( p \) and \( q \) first, then compare \( p+q \) with 1.
2. The four regions are fixed by these three tests.

Answer: (a) outside triangle OAB but inside angle OBA

Explanation:
1. Here \( p=\dfrac34>0 \), \( q=-\dfrac12<0 \) and \( p+q=\dfrac14<1 \).
2. This is the case outside the triangle but inside angle OBA.

Teacher's Note:
1. Negative \( q \) puts the point on the opposite side of line OA from B.
2. Apply the rule given for the case \( p>0,q<0 \).

Answer: (d) \( \sin^{-1}\dfrac13 \)

Explanation:
1. The direction is \( (1,1,1) \) and the plane normal is \( (1,-1,-1) \).
2. \( \sin\phi=\dfrac{|(1,1,1)\cdot(1,-1,-1)|}{\sqrt3\sqrt3}=\dfrac13 \).
3. The angle is \( \sin^{-1}\dfrac13 \).

Teacher's Note:
1. The angle between a line and a plane uses the sine with the normal.
2. Do not use the cosine of the direction-normal angle.

Answer: (c) \( \dfrac1{\sqrt2} \)

Explanation:
1. Volume \( =\dfrac16\,(AB)(CD)\,d\sin\theta \) with opposite edges perpendicular (\( \theta=90^\circ \)) and lengths 1: \( \dfrac d6=\dfrac1{6\sqrt2} \).
2. So \( d=\dfrac1{\sqrt2} \).

Teacher's Note:
1. Opposite edges of a regular tetrahedron are perpendicular.
2. Use the volume formula in terms of opposite edges.

Answer: (d) \( 2\hat i+8\hat j-3\hat k \)

Explanation:
1. \( P=(5+3s,\,7-s,\,-2+s) \), \( Q=(-3-3t,\,3+2t,\,6+4t) \). Parallel to \( (2,7,-5) \) gives two equations whose solution (checked with sympy) is \( s=-1,\ t=-1 \).
2. \( P=(2,8,-3) \) and \( Q=(0,1,2) \), and \( \overrightarrow{PQ}=(-2,-7,5) \), which is parallel to \( (2,7,-5) \).

Teacher's Note:
1. Write both points with parameters and impose parallelism through proportional components.
2. Verify by computing \( \overrightarrow{PQ} \) at the end.

Answer: (d) \( \sqrt{78} \)

Explanation:
1. From the parameters \( s=t=-1 \), \( P=(2,8,-3) \) and \( Q=(0,1,2) \).
2. \( PQ=\sqrt{4+49+25}=\sqrt{78} \).

Teacher's Note:
1. Find the parameter values first, then use the distance formula.
2. Check that \( \overrightarrow{PQ} \) is a multiple of the given direction.

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

Explanation:
1. \( D=(2,\tfrac43,\tfrac83) \) and \( E=(\tfrac73,-\tfrac13,\tfrac{13}3) \). Lines CD and BE meet at \( F=(2,0,4) \) (parameters 3/7 and 6/7).
2. Normal to the plane: \( \overrightarrow{AB}\times\overrightarrow{AC}=(0,7,7) \), unit normal \( \dfrac{(0,1,1)}{\sqrt2} \).
3. \( \overrightarrow{FP}=\sqrt2\cdot\dfrac{(0,1,1)}{\sqrt2}=\hat j+\hat k \).

Teacher's Note:
1. Find F from the intersection of two cevians.
2. FP is along the unit normal scaled by the distance.

Answer: (a) \( \dfrac73 \) cubic units

Explanation:
1. The base area is \( \tfrac12|\overrightarrow{AB}\times\overrightarrow{AC}|=\tfrac12\cdot7\sqrt2 \) and the height is \( |\overrightarrow{FP}|=\sqrt2 \).
2. Volume \( =\tfrac13\cdot\dfrac{7\sqrt2}2\cdot\sqrt2=\dfrac73 \) cubic units.

Teacher's Note:
1. Volume of a pyramid is one third of base area times height.
2. The height is the length of the perpendicular \( FP \).

Answer: (b) \( 18 \)

Explanation:
1. The point moves on the circle \( (x+4)^2+(y-5)^2=1 \), and the distance from \( (-2,3) \) to its centre is \( 2\sqrt2 \).
2. \( p_1=(2\sqrt2+1)^2=9+4\sqrt2 \) and \( p_2=(2\sqrt2-1)^2=9-4\sqrt2 \).
3. \( p_1+p_2=18 \).

Teacher's Note:
1. The maximum and minimum squared distances come from the farthest and nearest points.
2. The irrational parts cancel in the sum.

Answer: (a) \( \tfrac12(2\vec a+\vec p+2\vec q) \)

Explanation:
1. \( B=\vec a+\vec p \), \( C=\vec a+\vec p+\vec q \), \( D=\vec a+\vec q \).
2. Midpoint of CD \( =\tfrac12(C+D)=\tfrac12(2\vec a+\vec p+2\vec q) \).

Teacher's Note:
1. Write each vertex by adding side vectors from A.
2. Midpoint is the average of the endpoints.

Answer: (d) \( \overrightarrow{OA} \)

Explanation:
1. \( \overrightarrow{OE}=\tfrac12(\vec c+\vec a) \), \( \overrightarrow{OF}=\tfrac12(\vec a+\vec b) \), \( \overrightarrow{DO}=-\tfrac12(\vec b+\vec c) \).
2. The sum is \( \tfrac12(\vec c+\vec a+\vec a+\vec b-\vec b-\vec c)=\vec a=\overrightarrow{OA} \).

Teacher's Note:
1. Write each midpoint as the average of two position vectors.
2. Terms cancel in pairs.

Answer: (d) \( 2 \)

Explanation:
1. \( |\vec a+\vec b+\vec c|^2=3|\vec a|^2+6|\vec a|^2\cdot\tfrac12=6|\vec a|^2=6 \), so \( |\vec a|=1 \).
2. \( 2|\vec a|=2 \).

Teacher's Note:
1. Expand the square of the sum using pairwise dot products.
2. Each dot product equals \( |\vec a|^2\cos60^\circ \).