JEE Mathematics Vector Algebra MCQs Set 15

Answer: (a) \( 2 \)

Explanation:
1. Coefficient determinant \( \begin{vmatrix}1&1&1\\1&\lambda&2\\-1&1&1\end{vmatrix}=(\lambda-2)-(1+2)+(1+\lambda)=2\lambda-4 \).
2. It must vanish, so \( \lambda=2 \).

Teacher's Note:
1. Dependence of combinations of independent vectors means the coefficient determinant is zero.
2. Expand along the first row.

Answer: (c) \( \tfrac12(5\hat i+6\hat j+2\hat k) \)

Explanation:
1. \( \overrightarrow{BA}=(-3,0,0) \) and \( \overrightarrow{BC}=(0,4,0) \) are perpendicular, so the angle at B is \( 90^\circ \).
2. The circumcentre is the midpoint of the hypotenuse AC: \( \dfrac{(1,1,1)+(4,5,1)}2=\tfrac12(5,6,2) \).

Teacher's Note:
1. In a right-angled triangle the circumcentre is the midpoint of the hypotenuse.
2. Test for a right angle using a dot product.

Answer: (c) \( \sqrt{33} \)

Explanation:
1. The median vector is \( \tfrac12(\overrightarrow{AB}+\overrightarrow{AC})=\tfrac12(8,-2,8)=(4,-1,4) \).
2. Length \( =\sqrt{16+1+16}=\sqrt{33} \).

Teacher's Note:
1. The median from A is half the sum of the two side vectors from A.
2. Do not forget the factor one half.

Answer: (d) \( 3\overrightarrow{GG'} \)

Explanation:
1. \( \overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}=(\vec a'+\vec b'+\vec c')-(\vec a+\vec b+\vec c) \).
2. \( =3\vec g'-3\vec g=3\overrightarrow{GG'} \).

Teacher's Note:
1. Write each displacement as head minus tail.
2. The sums of vertices are three times the centroids.

Answer: (d) \( \dfrac{\hat j+\hat k}{\sqrt2} \)

Explanation:
1. The bisector of the unit vectors \( \hat j \) and \( \hat k \) is along \( \hat j+\hat k \).
2. Normalising: \( \dfrac{\hat j+\hat k}{\sqrt2} \).

Teacher's Note:
1. The internal bisector of two unit vectors is along their sum.
2. Divide by the magnitude to normalise.

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

Explanation:
1. In a regular hexagon, \( \overrightarrow{AD}=2\overrightarrow{BC}=2\vec b \) and \( \overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD} \).
2. \( 2\vec b=\vec a+\vec b+\overrightarrow{CD} \), so \( \overrightarrow{CD}=\vec b-\vec a \).

Teacher's Note:
1. The long diagonal of a regular hexagon is parallel to a side and twice as long.
2. Use the triangle law along A to D.

Answer: (c) \( 24 \)

Explanation:
1. With vertices at \( A(1,0),B(\tfrac12,\tfrac{\sqrt3}2),C(-\tfrac12,\tfrac{\sqrt3}2),D(-1,0),E(-\tfrac12,-\tfrac{\sqrt3}2),F(\tfrac12,-\tfrac{\sqrt3}2) \): \( \overrightarrow{AD}=(-2,0)=2\overrightarrow{BC} \), so \( x=2 \).
2. \( \overrightarrow{CF}=(1,-\sqrt3)=2\overrightarrow{BA} \), so \( y=2 \).
3. \( (x+y)^2+8=16+8=24 \).

Teacher's Note:
1. Use coordinates on the unit circle for a regular hexagon.
2. Compare vectors component by component.

Answer: (a) \( (2014,\,2015,\,2016) \)

Explanation:
1. A point of the line is \( (2+t,\,3+t,\,4+t) \).
2. The coordinates differ by 1 each, which is true for \( (2014,2015,2016) \) at \( t=2012 \).
3. The other options do not have \( y-x=1 \) and \( z-y=1 \).

Teacher's Note:
1. Points on the line have coordinates that increase together with the parameter.
2. Check the differences between coordinates.

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

Explanation:
1. Equate: \( \vec a+s\vec b+s\vec c=\vec b+t\vec a+t\vec c \).
2. Independence gives \( 1=t \), \( s=1 \) and \( s=t \).
3. The point is \( \vec a+\vec b+\vec c \).

Teacher's Note:
1. Equate the coefficients of the independent vectors.
2. A single consistent solution gives the intersection point.

Answer: (d) \( a=1,\ c=0 \) and \( b \) is any real number

Explanation:
1. The line through \( (1,1,0) \) and \( (1,-1,0) \) is \( x=1,\ z=0 \) with y free.
2. R lies on it only if \( a=1 \), \( c=0 \), and \( b \) may be anything.

Teacher's Note:
1. Write the line through the first two points in coordinates.
2. The third point must satisfy every coordinate condition.

Answer: (d) \( 8 \)

Explanation:
1. The slope between the first two points is \( \dfrac{-5-3}{12-10}=-4 \).
2. For the third point: \( \dfrac{11-3}{\lambda-10}=-4 \), so \( \lambda-10=-2 \).
3. \( \lambda=8 \).

Teacher's Note:
1. Collinear points have equal slopes between pairs.
2. Use the first point as the common reference.

Answer: (b) \( \pm\vec c \)

Explanation:
1. \( |\vec a+\vec b|=|(6,-2,-4)|=\sqrt{56} \), so \( |\vec r|=\dfrac{\sqrt{56}}2=\sqrt{14} \).
2. \( |\vec c|=\sqrt{14} \), so \( \vec r=\pm\vec c \).

Teacher's Note:
1. Compare the required length with the length of \( \vec c \).
2. Collinear vectors differ only by a scalar factor.

Answer: (a) \( 9 \)

Explanation:
1. The terminal points are \( (3,4) \), \( (6,7) \) and \( (9,1+\lambda) \).
2. The slope between the first two is 1, so \( \dfrac{1+\lambda-4}{9-3}=1 \).
3. \( \lambda=9 \).

Teacher's Note:
1. Add each vector to the initial point to get the terminal points.
2. Then use equal slopes.

Answer: (c) never coplanar

Explanation:
1. The determinant of the three vectors is \( a(15a^2+31a+37) \) (expanded with sympy).
2. The quadratic \( 15a^2+31a+37 \) has discriminant \( 31^2-4\cdot15\cdot37<0 \), so it is never zero, and \( a\ne0 \).
3. The determinant is never zero, so the vectors are never coplanar.

Teacher's Note:
1. Use the determinant test for coplanarity.
2. Check the discriminant of the remaining quadratic factor.

Answer: (b) \( x=\dfrac{10}7,\ y=\dfrac47 \)

Explanation:
1. \( 2\vec u-\vec v=(2x+2y)\vec a+(4y-3x)\vec b \).
2. Comparing with \( 4\vec a-2\vec b \): \( 2x+2y=4 \) and \( 4y-3x=-2 \).
3. From the first, \( x=2-y \); then \( 4y-6+3y=-2 \), so \( y=\dfrac47 \) and \( x=\dfrac{10}7 \).

Teacher's Note:
1. Compare coefficients of the independent vectors \( \vec a \) and \( \vec b \).
2. Solve the two linear equations.

Answer: (c) \( (0,-4,-6) \)

Explanation:
1. The direction is \( (2,0,0)-(1,-2,-3)=(1,2,3) \). A point of the line is \( (2,0,0)+t(1,2,3) \).
2. For \( x=0 \), \( t=-2 \), giving \( (0,-4,-6) \).

Teacher's Note:
1. Find the line through the two points and set the x-coordinate to 0.
2. Test each option against the direction ratios.

Answer: (b) \( \dfrac27 \)

Explanation:
1. Take O as origin: \( M=\dfrac{\vec a+\vec b}2 \), \( C=\dfrac13M=\dfrac{\vec a+\vec b}6 \), \( X=\dfrac23\vec b \).
2. A point on XC: \( X+t(C-X)=\dfrac t6\vec a+\left(\dfrac23-\dfrac t2\right)\vec b \).
3. On OA the coefficient of \( \vec b \) is zero: \( t=\dfrac43 \), so \( Y=\dfrac29\vec a \).
4. \( OY=\dfrac29OA \), \( YA=\dfrac79OA \), ratio \( \dfrac27 \).

Teacher's Note:
1. Take the vertex O as origin and write each point in terms of \( \vec a \) and \( \vec b \).
2. A point on OA has no \( \vec b \) component.

Answer: (b) \( \dfrac83 \)

Explanation:
1. Take A as origin, \( \overrightarrow{AB}=\vec b \), \( \overrightarrow{AC}=\vec c \). Then \( D=\dfrac{\vec b+2\vec c}3 \) and \( E=\dfrac34\vec c \).
2. P on AD: \( t\,D \); P on BE: \( \vec b+s(E-\vec b) \). Equating coefficients gives \( s=\dfrac8{11} \) (and \( t=\dfrac9{11} \)).
3. \( \dfrac{BP}{PE}=\dfrac{s}{1-s}=\dfrac{8/11}{3/11}=\dfrac83 \).

Teacher's Note:
1. Write P in two ways (once on each line) and equate coefficients.
2. The ratio on BE is \( s:(1-s) \).

Answer: (d) \( 0 \)

Explanation:
1. Equating components: \( 2l+m-2n=3 \), \( -l+3m+n=2 \), \( l-2m-3n=-5 \).
2. Solving (checked with sympy): \( l=3,\ m=1,\ n=2 \).
3. \( l-m-n=3-1-2=0 \).

Teacher's Note:
1. Equate the \( \hat i \), \( \hat j \), \( \hat k \) components to get three linear equations.
2. Check the solution in all three equations.

Answer: (b) \( (2\sqrt5,\ \sqrt5,\ -5) \)

Explanation:
1. The z-component is \( 5\sqrt2\cos135^\circ=-5 \).
2. With \( x=2y \): \( 4y^2+y^2+25=50 \), so \( y^2=5 \) and \( y=\sqrt5 \), \( x=2\sqrt5 \).
3. The vector is \( (2\sqrt5,\sqrt5,-5) \).

Teacher's Note:
1. The z-component is magnitude times the cosine of the angle with the z-axis.
2. Use the magnitude to find y.

Answer: (d) \( (2,8,-4) \)

Explanation:
1. A rotation about the X-axis leaves the x-component unchanged: \( A_1=2 \).
2. For \( \alpha=\dfrac\pi2 \): \( A_2=a_3=8 \) and \( A_3=-a_2=-4 \).
3. The components are \( (2,8,-4) \).

Teacher's Note:
1. A rotation about the X-axis does not change the x-component.
2. Substitute \( \cos\dfrac\pi2=0 \) and \( \sin\dfrac\pi2=1 \).

Answer: (c) circumcentre

Explanation:
1. \( |\vec a-\vec d| \) is the distance DA, and similarly for DB and DC.
2. \( DA=DB=DC \) means D is equidistant from the three vertices, which defines the circumcentre.

Teacher's Note:
1. Equal distances from all three vertices define the circumcentre.
2. The incentre is equidistant from the sides, not the vertices.

Answer: (a) \( \lambda+1 \)

Explanation:
1. \( \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AB}+\lambda\overrightarrow{AD} \). \( \overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}=-\overrightarrow{AB}+\overrightarrow{AD} \).
2. The sum is \( (\lambda+1)\overrightarrow{AD} \). Answer: \( p=\lambda+1 \).

Teacher's Note:
1. Express both diagonals using \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \).
2. The \( \overrightarrow{AB} \) terms cancel.