JEE Mathematics Vector Algebra MCQs Set 09

Answer: (b) \( \frac{5\pi}{6} \)

Explanation:
1. Expand: \( \vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c \).
2. Since \( \vec b,\vec c \) are non-parallel, compare coefficients with \( \frac{\sqrt3}{2}\vec b+\frac{\sqrt3}{2}\vec c \): \( \vec a\cdot\vec c=\frac{\sqrt3}{2} \) and \( \vec a\cdot\vec b=-\frac{\sqrt3}{2} \).
3. All are unit vectors, so \( \cos\theta=-\frac{\sqrt3}{2} \) and \( \theta=\frac{5\pi}{6} \).

Teacher's Note:
1. Use the vector triple product expansion and compare coefficients of independent vectors.
2. Check that the dot product signs give an obtuse angle.

Answer: (c) 2

Explanation:
1. \( \vec a\times\vec b=(2,-2,1) \), so \( |\vec a\times\vec b|=3 \).
2. \( |(\vec a\times\vec b)\times\vec c|=|\vec a\times\vec b||\vec c|\sin30^\circ=\frac32|\vec c|=3 \), hence \( |\vec c|=2 \).
3. \( |\vec a|=3 \), and \( |\vec c-\vec a|^2=|\vec c|^2+|\vec a|^2-2\vec a\cdot\vec c \) gives \( 9=4+9-2\vec a\cdot\vec c \), so \( \vec a\cdot\vec c=2 \).

Teacher's Note:
1. Find the magnitude of the cross product first.
2. Use the magnitude formula for a vector product to get |c|.

Answer: (a) 6

Explanation:
1. Projection vector of b on a is \( \frac{\vec a\cdot\vec b}{|\vec a|^2}\vec a \) equating to \( \vec a \) gives \( \vec a\cdot\vec b=|\vec a|^2=4 \), so \( b_1+b_2+2=4 \).
2. \( (\vec a+\vec b)\cdot\vec c=0 \): \( 5(1+b_1)+(1+b_2)+4=0 \), i.e. \( 5b_1+b_2=-10 \).
3. Solving: \( b_1=-3,\ b_2=5 \).
4. \( |\vec b|=\sqrt{9+25+2}=6 \).

Teacher's Note:
1. Projection vector equal to a means a.b = |a|^2.
2. Two linear equations in b1 and b2.

Answer: (c) \( \frac{\sqrt3}{2} \)

Explanation:
1. The normal is \( (1,1,1)\times(1,2,3)=(1,-2,1) \), with magnitude \( \sqrt6 \).
2. Projection \( =\frac{|(2,3,1)\cdot(1,-2,1)|}{\sqrt6}=\frac{|2-6+1|}{\sqrt6}=\frac{3}{\sqrt6}=\frac{\sqrt6}{2} \).

Teacher's Note:
1. The vector perpendicular to the plane is the cross product.
2. Projection magnitude = |dot product| / |normal|.

Answer: (d) \( \frac{1}{\sqrt{15}} \)

Explanation:
1. \( M=(2,1,0) \) \( G \) is the centroid \( \frac{A+B+C}{3}=(2,4,2) \).
2. \( \cos\angle GOA=\frac{\vec{OG}\cdot\vec{OA}}{|OG||OA|}=\frac{6-2}{\sqrt{24}\sqrt{10}} \).
3. This is \( \frac{4}{\sqrt{240}}=\frac{1}{\sqrt{15}} \).

Teacher's Note:
1. BM is a median, so G is the centroid.
2. Use the dot product of OG and OA.

Answer: (d) \( \left(-\frac32,\,3\vec a\times\vec b\right) \)

Explanation:
1. Square the condition: \( 0=|\vec a+\vec b+\vec c|^2=3+2\lambda \), so \( \lambda=-\frac32 \).
2. Put \( \vec c=-\vec a-\vec b \): \( \vec b\times\vec c=-\vec b\times\vec a=\vec a\times\vec b \) and \( \vec c\times\vec a=-\vec b\times\vec a=\vec a\times\vec b \).
3. Hence \( \vec d=3\,\vec a\times\vec b \).

Teacher's Note:
1. Square the given vector sum.
2. Eliminate c using c = -a - b.

Answer: (b) \( \frac{\pi}{3} \)

Explanation:
1. \( \vec c\cdot\vec d=5|\hat a|^2-4\hat a\cdot\hat b+10\hat a\cdot\hat b-8|\hat b|^2=5+6\cos\theta-8=0 \).
2. So \( \cos\theta=\frac12 \) and \( \theta=\frac\pi3 \).

Teacher's Note:
1. Expand the dot product using unit magnitudes.
2. Perpendicular means zero dot product.

Answer: (d) \( -\vec q+\left(\frac{\vec p\cdot\vec q}{\vec p\cdot\vec p}\right)\vec p \)

Explanation:
1. Take A as origin: B has position \( \vec q \) and the line AD is along \( \vec p \). The foot F of the perpendicular from B on AD is \( \frac{\vec p\cdot\vec q}{\vec p\cdot\vec p}\vec p \).
2. The altitude vector from B to F is \( \vec{BF}=\vec F-\vec q=-\vec q+\frac{\vec p\cdot\vec q}{\vec p\cdot\vec p}\vec p \).

Teacher's Note:
1. The foot of the perpendicular is the projection of AB on AD.
2. BF = F - B.

Answer: (d) 1

Explanation:
1. Using \( [\vec a\times\vec b,\ \vec b\times\vec c,\ \vec c\times\vec a]=[\vec a\ \vec b\ \vec c]^{2} \) (the scalar triple product of the reciprocal-type set).
2. Check with \( \vec a=\hat i,\vec b=\hat j,\vec c=\hat k \): the three cross products are \( \hat k,\hat i,\hat j \) with triple product 1, and \( [\hat i\ \hat j\ \hat k]^2=1 \).
3. Hence \( \lambda=1 \).

Teacher's Note:
1. Test the identity on the unit vectors i, j, k.
2. The identity holds for all vectors.

Answer: (b) \( \frac{2\sqrt2}{3} \)

Explanation:
1. \( (\vec a\times\vec b)\times\vec c=(\vec a\cdot\vec c)\vec b-(\vec b\cdot\vec c)\vec a \).
2. Compare with \( \frac13|\vec b||\vec c|\vec a \): since \( \vec a,\vec b \) are independent, \( \vec a\cdot\vec c=0 \) and \( -\vec b\cdot\vec c=\frac13|\vec b||\vec c| \).
3. So \( \cos\theta=-\frac13 \) and \( \sin\theta=\sqrt{1-\frac19}=\frac{2\sqrt2}{3} \) (positive for an angle between 0 and pi).

Teacher's Note:
1. Expand the triple product and match coefficients.
2. sin theta is non-negative for angles in [0, pi].

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

Explanation:
1. \( \vec a\cdot\vec c=\lambda-1+2\mu=0 \) and \( \vec b\cdot\vec c=2\lambda+4+\mu=0 \).
2. From the second, \( \mu=-2\lambda-4 \); the first gives \( \lambda-1-4\lambda-8=0 \), so \( \lambda=-3,\ \mu=2 \).
3. (Also \( \vec a\cdot\vec b=2-4+2=0 \), as required.)

Teacher's Note:
1. Mutually orthogonal means every pair has zero dot product.
2. Use the two conditions involving c.

Answer: (a) -5

Explanation:
1. Both vectors are unit vectors and \( \vec a\cdot\vec b=\frac{1}{7\sqrt{10}}(6-6)=0 \).
2. Let \( \vec u=\vec a\times\vec b \). Then \( \vec a,\vec b,\vec u \) are mutually orthogonal unit vectors forming a right-handed triad, so \( \vec u\times\vec a=\vec b \) and \( \vec u\times\vec b=-\vec a \).
3. So the bracket is \( \vec b-2\vec a \) and the product is \( (2\vec a-\vec b)\cdot(\vec b-2\vec a)=-|2\vec a-\vec b|^2=-(4+1)=-5 \).

Teacher's Note:
1. a and b are orthogonal unit vectors, so a, b, a x b form an orthonormal triad.
2. Use u x a = b and u x b = -a.

Answer: (b) -2

Explanation:
1. Coplanarity: \( \begin{vmatrix}p&1&1\\1&q&1\\1&1&r\end{vmatrix}=0 \).
2. Expanding: \( pqr-p-q-r+2=0 \).
3. Hence \( pqr-(p+q+r)=-2 \).

Teacher's Note:
1. Set the scalar triple product to zero.
2. Expand the 3 x 3 determinant.

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

Explanation:
1. Since \( |\vec w|=1 \) and \( \vec w\cdot(\hat u\times\vec u)=1=|\hat u\times\vec u| \), we get \( \hat u\times\vec u=\vec w \). This requires \( \vec w\perp\hat u \), i.e. \( u_1+u_2+2u_3=0 \).
2. Then \( \vec u \) can be changed by any multiple of \( \hat u \) without changing the cross product: infinitely many choices, so (ii) holds.
3. (iii) \( u_3=0 \) gives \( u_1=-u_2 \): true. (iv) \( u_2=0 \) gives \( u_1=-2u_3 \), i.e. \( |u_1|=2|u_3| \), so \( 2|u_1|=|u_3| \) is false.

Teacher's Note:
1. The only way a unit-vector dot product with a unit vector equals 1 is if they coincide.
2. The cross product is unchanged when a multiple of u-hat is added to u.

Answer: (d) \( -\frac32 \)

Explanation:
1. Since \( P+Q=\pi-R \) etc., the expression equals \( -(\cos P+\cos Q+\cos R) \).
2. In any triangle \( \cos P+\cos Q+\cos R\le\frac32 \) (equality for the equilateral triangle).
3. Hence the minimum is \( -\frac32 \).

Teacher's Note:
1. Use A + B + C = pi to rewrite each cosine.
2. Recall the maximum of the sum of cosines of a triangle.

Answer: (a) \( \sin(P+Q) \)

Explanation:
1. The angle between the directions of \( \vec{QR} \) and \( \vec{RP} \) is the exterior angle at R, namely \( \pi-R \).
2. So \( |\overrightarrow{OX}\times\overrightarrow{OY}|=\sin(\pi-R)=\sin R=\sin(P+Q) \).

Teacher's Note:
1. The angle between two head-to-tail side vectors is an exterior angle.
2. sin(pi - R) = sin R.

Answer: (d) \( \frac12 \)

Explanation:
1. \( \vec p=(\tfrac12,-\tfrac12,-\tfrac12),\ \vec q=(-\tfrac12,\tfrac12,-\tfrac12),\ \vec r=(-\tfrac12,-\tfrac12,\tfrac12),\ \vec t=(\tfrac12,\tfrac12,\tfrac12) \).
2. \( \vec p\times\vec q=(\tfrac12,\tfrac12,0) \) and \( \vec r\times\vec t=(-\tfrac12,\tfrac12,0) \).
3. Their cross product is \( (0,0,\tfrac14+\tfrac14)=(0,0,\tfrac12) \), of magnitude \( \frac12 \).

Teacher's Note:
1. Write the four vectors from the centre explicitly.
2. Evaluate the two inner cross products first.

Answer: (b) \( \frac34 \)

Explanation:
1. \( A=(1,0,0),\ B=(\tfrac12,\tfrac12,0),\ C=(\tfrac13,\tfrac13,\tfrac13) \).
2. \( \vec{AB}=(-\tfrac12,\tfrac12,0),\ \vec{AC}=(-\tfrac23,\tfrac13,\tfrac13) \), and \( \vec{AB}\times\vec{AC}=(\tfrac16,\tfrac16,\tfrac16) \) with magnitude \( \frac{\sqrt3}{6} \).
3. So \( \Delta=\frac{\sqrt3}{12} \), \( 6\Delta=\frac{\sqrt3}{2} \) and \( (6\Delta)^2=\frac34 \).

Teacher's Note:
1. Find the three points of intersection.
2. Area = half the magnitude of the cross product.

Answer: (a) \( 15\sqrt5 \)

Explanation:
1. Treat \( (f,g,h) \) and \( (3,4,10) \) as vectors; by the Cauchy-Schwarz inequality \( U\le\sqrt{f^2+g^2+h^2}\sqrt{9+16+100} \).
2. So \( U\le3\sqrt{125}=15\sqrt5 \), attained when \( (f,g,h) \) is parallel to \( (3,4,10) \) with length 3.

Teacher's Note:
1. Use the dot product bound a.b <= |a||b|.
2. |(3,4,10)|^2 = 125.

Answer: (b) 19

Explanation:
1. Each term \( \overrightarrow{OA_i}\times\overrightarrow{OA_{i+1}} \) has the same magnitude \( R^2\sin18^\circ \) and the same direction (perpendicular to the plane) when the vertices are taken in order.
2. The right-hand vector \( \overrightarrow{OA_2}\times\overrightarrow{OA_1} \) has the same magnitude but opposite direction.
3. So the sum of 19 terms equals \( -19\left(\overrightarrow{OA_2}\times\overrightarrow{OA_1}\right) \), i.e. \( \lambda=-19 \) and \( |\lambda|=19 \).

Teacher's Note:
1. All consecutive cross products are equal for a regular polygon.
2. Reversing the order flips the sign.

Answer: (c) 72

Explanation:
1. Write the circumcentre as \( \vec r=x\,(\vec b\times\vec c)+y\,(\vec c\times\vec a)+z\,(\vec a\times\vec b) \) and use \( |\vec r|^2=|\vec r-\vec a|^2 \) etc., i.e. \( 2\vec r\cdot\vec a=a^2 \), \( 2\vec r\cdot\vec b=b^2 \), \( 2\vec r\cdot\vec c=c^2 \).
2. Dotting with \( \vec a \): \( \vec r\cdot\vec a=x[\vec a\,\vec b\,\vec c] \), so \( x=\frac{a^2}{2[\vec a\,\vec b\,\vec c]} \); similarly for y and z.
3. Hence the denominator is \( \alpha=2[\vec a\ \vec b\ \vec c]=72 \).

Teacher's Note:
1. The circumcentre is equidistant from O, A, B and C.
2. Use the reciprocal basis (b x c, c x a, a x b).

Answer: (b) 36

Explanation:
1. Points \( C=(7-3p,\ 6+2p,\ 2+4p) \) and \( D=(5+2q,\ 3+q,\ 4+3q) \).
2. \( \overrightarrow{CD}=D-C=m(2,-2,-1) \) for some \( m \). Solving the three equations gives \( p=2,\ q=1,\ m=-3 \), so \( C=(1,10,10) \), \( D=(7,4,7) \).
3. \( \overrightarrow{CD}=(6,-6,-3) \) with magnitude \( 9 \), so \( |4\overrightarrow{CD}|=36 \).

Teacher's Note:
1. Write general points on the two lines.
2. CD must be parallel to the given direction.

Answer: (a) \( p=1 \text{ or } p=-\frac13 \)

Explanation:
1. Rotation of axes preserves the magnitude of the vector: \( (2p)^2+1^2=(p+1)^2+1^2 \).
2. So \( 4p^2=p^2+2p+1 \), i.e. \( 3p^2-2p-1=0 \), which factorises as \( (3p+1)(p-1)=0 \).
3. Hence \( p=1 \) or \( p=-\frac13 \).

Teacher's Note:
1. A rotation does not change length.
2. Solve the resulting quadratic.

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

Explanation:
1. Let \( Q \) be the matrix with rows \( (a,b) \) and \( (c,d) \). Its rows are orthogonal and of equal length \( r \), so \( a^2+b^2=c^2+d^2=r^2 \) and \( ac+bd=0 \).
2. Hence \( QQ^T=r^2I \), and therefore \( Q^TQ=r^2I \) too: the columns are also orthogonal with length \( r \).
3. So \( a^2+c^2=b^2+d^2=r^2 \) and \( ab+cd=0 \): (i), (ii), (iii) are true and (iv) is false.

Teacher's Note:
1. Write the condition as a matrix equation with rows z1 and z2.
2. A square matrix with orthogonal rows of equal length also has orthogonal columns of that length.