JEE Mathematics Vectors MCQs Set 31

Answer: (a) \( \frac{u\cos\alpha}{f} \)

Explanation:
1. Take the first particle's direction as the \( x \)-axis. Relative velocity of the second w.r.t. the first is \( \vec v_{21}=\vec ft-\vec u \).
2. \( |\vec v_{21}|^2=f^2t^2-2uft\cos\alpha+u^2 \) is least when \( t=\frac{u\cos\alpha}f \).

Teacher's Note:
1. Minimise the square of the speed, a quadratic in \( t \).
2. The vertex of \( at^2+bt+c \) is at \( -\frac b{2a} \).

Answer: (b) \( \frac{-4\hat i+3\hat j+\hat k}{\sqrt{26}} \)

Explanation:
1. A vector perpendicular to both is \( (\hat i+\hat j+\hat k)\times(2\hat i+3\hat j-\hat k)=-4\hat i+3\hat j+\hat k \).
2. Its magnitude is \( \sqrt{16+9+1}=\sqrt{26} \), so the unit vector is \( \frac{-4\hat i+3\hat j+\hat k}{\sqrt{26}} \).

Teacher's Note:
1. The cross product is perpendicular to both vectors.
2. Divide by its magnitude to get a unit vector.

Answer: (b) Geometric mean of \( \alpha \) and \( \beta \)

Explanation:
1. Coplanar vectors have zero scalar triple product: \( \begin{vmatrix}\alpha&\alpha&\gamma\\1&0&1\\\gamma&\gamma&\beta\end{vmatrix}=0 \).
2. Expanding: \( -\alpha\gamma-\alpha(\beta-\gamma)+\gamma^2=\gamma^2-\alpha\beta=0 \), so \( \gamma=\sqrt{\alpha\beta} \), the geometric mean.

Teacher's Note:
1. Three vectors are coplanar when their determinant vanishes.
2. \( \gamma^2=\alpha\beta \) is the definition of the geometric mean.

Answer: (c) \( \frac{(\vec a\cdot\vec b)(\vec b\times\vec a)}{|\vec b|^2} \)

Explanation:
1. \( \vec a_1=\frac{\vec a\cdot\vec b}{|\vec b|^2}\vec b \) and \( \vec a_2=\vec a-\vec a_1 \).
2. \( \vec a_1\times\vec a_2=\vec a_1\times\vec a=\frac{\vec a\cdot\vec b}{|\vec b|^2}(\vec b\times\vec a) \), since \( \vec a_1\times\vec a_1=\vec 0 \).

Teacher's Note:
1. The component along \( \vec b \) is the projection vector.
2. The cross product of parallel vectors is zero.

Answer: (c) 5

Explanation:
1. Put \( A \) at the origin with \( AB=3 \) along the \( x \)-axis and \( AC=4 \) along the \( y \)-axis. Since the moment about \( A \) is 0, the force passes through \( A \); let it make angle \( \varphi \) with \( AB \).
2. Moment about \( B \): \( F\cdot3\sin\varphi=9 \); about \( C \): \( F\cdot4\cos\varphi=16 \). So \( F\sin\varphi=3,\ F\cos\varphi=4 \) and \( F=5 \).

Teacher's Note:
1. Zero moment about a point means the line of action passes through it.
2. Moment equals force times perpendicular distance.

Answer: (c) \( |[\vec a\ \vec b\ \vec c]\vec r| \)

Explanation:
1. Since \( \vec r\cdot(\vec b-\vec a)=\vec r\cdot(\vec c-\vec a)=0 \), \( \vec r \) is parallel to \( \vec n=(\vec b-\vec a)\times(\vec c-\vec a) \), say \( \vec n=k\vec r \).
2. \( \vec n\cdot\vec a=[\vec a\ \vec b\ \vec c] \) and \( \vec n\cdot\vec a=k\vec r\cdot\vec a=\frac k2 \), so \( k=2[\vec a\ \vec b\ \vec c] \). Area \( =\frac12|\vec n|=|[\vec a\ \vec b\ \vec c]\vec r| \).

Teacher's Note:
1. The area of a triangle is half the magnitude of the cross product of two sides.
2. \( (\vec b-\vec a)\times(\vec c-\vec a)\cdot\vec a=[\vec a\ \vec b\ \vec c] \).

Answer: (d) \( \frac\pi2 \)

Explanation:
1. Let the first term be \( A \) and the common ratio \( R \). Then \( \log a=\log A+(p-1)\log R \), and similarly for \( b,c \).
2. The dot product is \( 2\log R\left[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)\right]+2\log A\cdot0=0 \), since \( \sum p(q-r)=0 \) and \( \sum(q-r)=0 \). The vectors are perpendicular: \( \frac\pi2 \).

Teacher's Note:
1. Write the logs of the terms as linear functions of \( p,q,r \).
2. A zero dot product means the vectors are at \( 90^\circ \).

Answer: (c) \( 5\sqrt3 \) square units

Explanation:
1. Area of a parallelogram \( =\frac12|\vec d_1\times\vec d_2| \).
2. \( (3,1,-2)\times(1,-3,4)=(4-6,\,-2-12,\,-9-1)=(-2,-14,-10) \), magnitude \( \sqrt{300}=10\sqrt3 \). Area \( =5\sqrt3 \).

Teacher's Note:
1. The diagonals' cross product is twice the area.
2. Simplify \( \sqrt{300}=10\sqrt3 \).

Answer: (a) 23 units

Explanation:
1. Moment \( =\vec r\times\vec F=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-2&-3\\3&4&-3\end{vmatrix}=18\hat i-3\hat j+14\hat k \).
2. Magnitude \( =\sqrt{324+9+196}=\sqrt{529}=23 \).

Teacher's Note:
1. The magnitude of the moment is the length of \( \vec r\times\vec F \).
2. \( 23^2=529 \).

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

Explanation:
1. \( \vec a\cdot\vec b=\frac12-\frac12+1=1 \) and \( |\vec a|=|\vec b|=\sqrt{\frac12+\frac12+1}=\sqrt2 \).
2. \( \cos\theta=\frac1{2}\Rightarrow\theta=60^\circ \), which is acute, so it is the interior acute angle.

Teacher's Note:
1. Use \( \cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|} \).
2. The other interior angle is \( 120^\circ \).

Answer: (a) 0

Explanation:
1. \( (\vec c\times\vec a)\cdot\vec b=[\vec c\ \vec a\ \vec b]=[\vec a\ \vec b\ \vec c] \), so the first term is 1.
2. \( \vec b\cdot(\vec a\times\vec c)=[\vec b\ \vec a\ \vec c]=-[\vec a\ \vec b\ \vec c] \) and \( \vec c\cdot(\vec a\times\vec b)=[\vec a\ \vec b\ \vec c] \), so the second term is \( -1 \). Sum \( =0 \).

Teacher's Note:
1. A scalar triple product is unchanged by cyclic shifts and changes sign on swapping two vectors.
2. Reduce everything to \( [\vec a\ \vec b\ \vec c] \).

Answer: (c) \( 3\vec p^{\,2} \)

Explanation:
1. \( |\vec p\times\hat i|^2=|\vec p|^2-(\vec p\cdot\hat i)^2 \), and similarly for \( \hat j,\hat k \).
2. Sum \( =3|\vec p|^2-|\vec p|^2=2|\vec p|^2 \), so \( \frac32\cdot2|\vec p|^2=3\vec p^{\,2} \).

Teacher's Note:
1. Use \( |\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2 \).
2. \( \sum(\vec p\cdot\hat i)^2=|\vec p|^2 \).

Answer: (c) \( \hat i+2\hat j+4\hat k \)

Explanation:
1. \( \vec r=2\hat i-\hat j \), \( \vec F=2\hat i+\hat j-\hat k \). Moment \( =\vec r\times\vec F=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-1&0\\2&1&-1\end{vmatrix} \).
2. \( =\hat i(1-0)-\hat j(-2-0)+\hat k(2+2)=\hat i+2\hat j+4\hat k \).

Teacher's Note:
1. Moment about the origin is \( \vec r\times\vec F \).
2. Watch the sign of the \( \hat j \) term in the determinant.

Answer: (b) 6

Explanation:
1. The three conditions give \( \vec a\cdot\vec b+\vec a\cdot\vec c=0,\ \vec b\cdot\vec c+\vec b\cdot\vec a=0,\ \vec c\cdot\vec a+\vec c\cdot\vec b=0 \). Adding pairs shows all three dot products are 0.
2. \( |\vec a+\vec b+\vec c|^2=16+16+4=36 \), so \( |\vec a+\vec b+\vec c|=6 \).

Teacher's Note:
1. Solve for each pairwise dot product from the linear system.
2. Then expand the square of the sum.

Answer: (b) \( -2\hat i+6\hat j-10\hat k \)

Explanation:
1. Resultant force \( \vec F=(1+2-1)\hat i+(2+3-1)\hat j+(-3+4+1)\hat k=2\hat i+4\hat j+2\hat k \). \( \vec r=\overrightarrow{AP}=(0,1,2)-(1,-2,0)=(-1,3,2) \).
2. \( \vec r\times\vec F=\begin{vmatrix}\hat i&\hat j&\hat k\\-1&3&2\\2&4&2\end{vmatrix}=-2\hat i+6\hat j-10\hat k \).

Teacher's Note:
1. The moment of several forces at a point is the moment of their resultant.
2. \( \vec r \) goes from the moment centre to the point of application.

Answer: (b) \( \frac73 \)

Explanation:
1. Projection of \( \vec a \) on \( \vec b \) \( =\frac{\vec a\cdot\vec b}{|\vec b|}=\frac{-4-6-6}{3}=-\frac{16}3 \).
2. Projection of \( \vec b \) on \( \vec a \) \( =\frac{\vec a\cdot\vec b}{|\vec a|}=\frac{-16}7 \). Ratio \( =\frac{|\vec a|}{|\vec b|}=\frac73 \).

Teacher's Note:
1. Both projections share the numerator \( \vec a\cdot\vec b \).
2. So the ratio is \( \frac{|\vec a|}{|\vec b|} \).

Answer: (c) orthocentre of \( \triangle ABC \)

Explanation:
1. \( (\vec a-\vec d)\cdot(\vec b-\vec c)=0 \) means \( DA\perp BC \); \( (\vec b-\vec d)\cdot(\vec c-\vec a)=0 \) means \( DB\perp CA \).
2. Two altitudes meet at \( D \), so \( D \) is the orthocentre.

Teacher's Note:
1. A zero dot product means perpendicular lines.
2. The orthocentre is the intersection of the altitudes.

Answer: (c) 3

Explanation:
1. Using the reciprocal system, \( \vec a\cdot\vec p=1 \) while \( \vec b\cdot\vec p=\vec c\cdot\vec p=0 \). So \( (\vec a-\vec b-\vec c)\cdot\vec p=1 \).
2. Likewise \( (\vec b-\vec c-\vec a)\cdot\vec q=1 \) and \( (\vec c-\vec a-\vec b)\cdot\vec r=1 \). The sum is \( 3 \).

Teacher's Note:
1. Use \( \vec a\cdot\vec p=1 \) and \( \vec b\cdot\vec p=\vec c\cdot\vec p=0 \).
2. Each bracket has the matching vector with coefficient 1.

Answer: (c) \( a_1+a_2+\ldots+a_n=0 \)

Explanation:
1. Shift the origin by \( \vec t \): the new position vectors are \( \vec r_i-\vec t \).
2. \( \sum a_i(\vec r_i-\vec t)=\sum a_i\vec r_i-\vec t\sum a_i=\vec 0 \) for every \( \vec t \) only if \( \sum a_i=0 \).

Teacher's Note:
1. Changing the origin subtracts the same vector from every position vector.
2. The condition must hold for any shift.

Answer: (a) \( x=2,y=1 \)

Explanation:
1. \( \vec u-\vec v=(2x-2y)\vec\alpha+(y-3x)\vec\beta=2\vec\alpha-5\vec\beta \).
2. By non-collinearity: \( x-y=1 \) and \( y-3x=-5\Rightarrow x=2,\ y=1 \).

Teacher's Note:
1. Compare coefficients of non-collinear vectors.
2. Solve the two linear equations.

Answer: (d) circle

Explanation:
1. Orthogonal: \( 1\cdot1+(-2x)(3x)+(-3y)(2y)=0 \).
2. \( 1-6x^2-6y^2=0\Rightarrow x^2+y^2=\frac16 \), a circle.

Teacher's Note:
1. Perpendicular vectors have zero dot product.
2. \( x^2+y^2=r^2 \) is a circle.

Answer: (b) 50

Explanation:
1. The three brackets are the components of \( (a,b,c)\times(3,4,5)=(5b-4c,\,3c-5a,\,4a-3b) \), so the expression is \( |(a,b,c)\times(3,4,5)|^2 \).
2. By Lagrange's identity this equals \( (a^2+b^2+c^2)(9+16+25)-(3a+4b+5c)^2=50-(3a+4b+5c)^2\le50 \), with equality when \( 3a+4b+5c=0 \). The maximum is 50.

Teacher's Note:
1. Spot the cross product pattern in the three squared brackets.
2. Lagrange's identity: \( |\vec u\times\vec v|^2=|\vec u|^2|\vec v|^2-(\vec u\cdot\vec v)^2 \).

Answer: (c) \( \vec a-\vec b+\vec c \)

Explanation:
1. In a parallelogram \( ABCD \), the diagonals bisect each other: \( \frac{\vec a+\vec c}2=\frac{\vec b+\vec d}2 \).
2. Hence \( \vec d=\vec a+\vec c-\vec b \).

Teacher's Note:
1. Midpoints of the diagonals \( AC \) and \( BD \) coincide.
2. Order the vertices as they appear in the name \( ABCD \).

Answer: (a) \( \vec\alpha \)

Explanation:
1. Write \( \vec\alpha=\alpha_1\hat i+\alpha_2\hat j+\alpha_3\hat k \). Then \( \vec\alpha\cdot\hat i=\alpha_1 \), etc.
2. The expression is \( \alpha_1\hat i+\alpha_2\hat j+\alpha_3\hat k=\vec\alpha \).

Teacher's Note:
1. This is the resolution of a vector into components.
2. Each term is the component along an axis.

Answer: (b) \( \pm2 \)

Explanation:
1. \( (\vec a+\vec b)\cdot(\vec a-\vec b)=|\vec a|^2-|\vec b|^2=0 \).
2. \( 1+4+9=9+1+\lambda^2\Rightarrow\lambda^2=4\Rightarrow\lambda=\pm2 \).

Teacher's Note:
1. The vectors \( \vec a+\vec b \) and \( \vec a-\vec b \) are perpendicular exactly when \( |\vec a|=|\vec b| \).
2. Take both roots.

Answer: (b) \( m<-2 \) or \( m>-\frac12 \)

Explanation:
1. The angle is acute when the dot product is positive: \( 2(1+m)+2m^2+3m=2m^2+5m+2>0 \).
2. \( (2m+1)(m+2)>0\Rightarrow m<-2 \) or \( m>-\frac12 \).

Teacher's Note:
1. Acute angle means \( \cos\theta>0 \).
2. Solve the quadratic inequality.

Answer: (b) \( (0,2) \)

Explanation:
1. \( \vec a=\hat i-3\hat j \), and \( \overrightarrow{AB}=\vec a=(1,-3) \).
2. \( B=A+\overrightarrow{AB}=(-1+1,\,5-3)=(0,2) \).

Teacher's Note:
1. The position vector of the point \( (1,-3) \) is \( \hat i-3\hat j \).
2. Add the displacement to the coordinates of \( A \).

Answer: (b) 1

Explanation:
1. Three points are collinear if \( \vec r_3=\lambda\vec r_1+(1-\lambda)\vec r_2 \) (a point dividing the line joining \( \vec r_1,\vec r_2 \)).
2. Comparing with \( \vec r_3=m\vec r_1+n\vec r_2 \): \( m+n=1 \).

Teacher's Note:
1. The coefficients of an affine combination add to 1.
2. This is the section formula in disguise.

Answer: (a) \( \frac{2\pi}3<\alpha<\frac{4\pi}3 \)

Explanation:
1. \( |\vec p+\vec q|^2=2+2\cos\alpha<1\Rightarrow\cos\alpha<-\frac12 \).
2. This holds for \( \frac{2\pi}3<\alpha<\frac{4\pi}3 \).

Teacher's Note:
1. Square the magnitude and use \( \vec p\cdot\vec q=\cos\alpha \).
2. Solve the trigonometric inequality on \( [0,2\pi] \).

Answer: (d) \( \lambda>1 \)

Explanation:
1. \( |\vec r_1|^2=\lambda^2+5 \) and \( |\vec r_2|^2=1+(2-\lambda)^2+4=\lambda^2-4\lambda+9 \).
2. \( \lambda^2+5>\lambda^2-4\lambda+9\Rightarrow4\lambda>4\Rightarrow\lambda>1 \).

Teacher's Note:
1. Compare squares of the magnitudes to avoid roots.
2. The \( \lambda^2 \) terms cancel.