JEE Mathematics Vector Algebra MCQs Set 06

Answer: (a) A, C and D only

Explanation:
1. N = (1, 2, 0), \( |OP|=\sqrt{14} \), \( |ON|=\sqrt5 \).
2. \( \cos\alpha=\frac{3}{\sqrt{14}},\ \sin\alpha=\frac{\sqrt5}{\sqrt{14}},\ \tan\alpha=\frac{\sqrt5}{3} \) so (C) is true.
3. \( \cos\beta=\frac1{\sqrt5},\ \sin\beta=\frac2{\sqrt5},\ \tan\beta=2 \) so (D) is true.
4. \( \sin\alpha\sin\beta=\frac{\sqrt5}{\sqrt{14}}\cdot\frac2{\sqrt5}=\frac2{\sqrt{14}} \) so (A) is true, while \( \cos\alpha\cos\beta=\frac{3}{\sqrt{70}} \) so (B) is false.

Teacher's Note:
1. Write the direction cosines of OP and ON from the coordinates.
2. Test each statement separately.

Answer: (c) \( 2\sqrt2,\ 2 \)

Explanation:
1. With angle \( \theta \): \( |\vec a+\vec b|=2\cos\frac\theta2 \) and \( |\vec a-\vec b|=2\sin\frac\theta2 \).
2. The sum is \( 2\left(\cos\frac\theta2+\sin\frac\theta2\right) \) with \( \frac\theta2\in[0,90^\circ] \).
3. Its maximum is \( 2\sqrt2 \) (at 45°) and its minimum is 2 (at 0° or 90°).

Teacher's Note:
1. For unit vectors |a ± b|² = 2 ± 2cos θ.
2. cos x + sin x ranges from 1 to √2 on [0°, 90°].

Answer: (b) 137

Explanation:
1. Take D as origin, \( \overrightarrow{DA}=\vec a,\ \overrightarrow{DB}=\vec b,\ \overrightarrow{DC}=\vec c \). The distance is \( d^2=\left|\frac{\vec a+\vec b}{2}-\frac{\vec c}{2}\right|^2 \).
2. Given lengths: \( |\vec a-\vec b|^2=1681,\ |\vec b-\vec c|^2=1296,\ |\vec c-\vec a|^2=49 \). Expanding, \( 4d^2=|\vec a|^2+|\vec b|^2+|\vec c|^2+2\vec a\cdot\vec b-2\vec a\cdot\vec c-2\vec b\cdot\vec c \).
3. From the three lengths, \( 2\vec a\cdot\vec b=324+729-1681=-628 \), \( 2\vec b\cdot\vec c=729+169-1296=-398 \), \( 2\vec a\cdot\vec c=324+169-49=444 \).
4. So \( 4d^2=1222-628-444+398=548 \) and \( d^2=137 \).

Teacher's Note:
1. The mid-point of AB minus the mid-point of CD is (a + b − c)/2 when D is the origin.
2. Get each dot product from the squared edge lengths via |x − y|² = |x|² + |y|² − 2x·y.

Answer: (b) \( \cos^{-1}\frac{3\sqrt7}{8} \)

Explanation:
1. Place A = (0, 0), C = (12, 0). B lies above the mid-point of AC with height \( \sqrt{64-36}=2\sqrt7 \): B = (6, 2√7).
2. E = A + (B − A)/4 = \( \left(\frac32,\frac{\sqrt7}{2}\right) \).
3. \( \overrightarrow{CE}=\left(-\frac{21}{2},\frac{\sqrt7}{2}\right),\ \overrightarrow{CA}=(-12,0) \).
4. \( \cos\theta=\frac{126}{12\sqrt{112}}=\frac{126}{48\sqrt7}=\frac{3\sqrt7}{8} \).

Teacher's Note:
1. Use coordinates for a triangle with given sides.
2. √112 = 4√7.

Answer: (c) 9

Explanation:
1. \( (\vec r-\vec c)\times\vec b=\vec 0 \), so \( \vec r=\vec c+t\vec b \).
2. \( \vec r\cdot\vec a=\vec c\cdot\vec a+t\,\vec b\cdot\vec a=-4+t=0\Rightarrow t=4 \).
3. \( \vec r=\vec c+4\vec b=(-3,6,3) \), so \( \vec r\cdot\vec b=3+6=9 \).

Teacher's Note:
1. Equal cross products with b mean the difference is parallel to b.
2. Fix the parameter with the dot-product condition.

Answer: (c) 3

Explanation:
1. With weights 1, 2, 3 at A, B, C, O is the weighted centre, so the area of AOC is proportional to the weight of the opposite vertex B.
2. \( \frac{[AOC]}{[ABC]}=\frac{2}{1+2+3}=\frac13 \).
3. So the ratio is 3.

Teacher's Note:
1. If αOA + βOB + γOC = 0, then [BOC] : [COA] : [AOB] = α : β : γ.
2. Cross products can also prove it: a×b = 3(b×c) = (3/2)(c×a) with the origin at O.

Answer: (a) 1

Explanation:
1. Three planes through the origin share a line exactly when the determinant of the coefficients is zero.
2. \( \begin{vmatrix}1&-c&-b\\c&-1&a\\b&a&-1\end{vmatrix}=1-a^2-b^2-c^2-2abc=0 \).
3. So \( a^2+b^2+c^2+2abc=1 \).

Teacher's Note:
1. Homogeneous planes always meet at the origin; a whole line of intersection needs a zero determinant.
2. Expand the determinant along the first row.

Answer: (b) 37

Explanation:
1. Magnitudes: |a| = 3, |b| = 7, |c| = 9. A vector \( \vec x \) equally inclined to them satisfies \( \frac{\vec x\cdot\vec a}{3}=\frac{\vec x\cdot\vec b}{7}=\frac{\vec x\cdot\vec c}{9} \), which gives \( \vec x=(3,5,7) \) with \( \vec x\cdot\vec a=27,\ \vec x\cdot\vec b=63,\ \vec x\cdot\vec c=81 \).
2. x is perpendicular to the plane, so \( \vec x\cdot(p\vec a-q\vec b)=0 \) and \( \vec x\cdot(q\vec b-r\vec c)=0 \): \( 27p=63q=81r \).
3. So \( p:q:r=\frac1{27}:\frac1{63}:\frac1{81}=21:9:7 \), the least integers are 21, 9, 7 and the sum is 37.

Teacher's Note:
1. Equal angles with three vectors means equal values of x·a/|a|.
2. Any two points on the plane differ by a vector perpendicular to x.

Answer: (b) 2

Explanation:
1. For a regular tetrahedron of edge L, the shortest distance between opposite edges is \( \frac{L}{\sqrt2} \).
2. With \( L=\sqrt8 \): \( \frac{\sqrt8}{\sqrt2}=2 \).

Teacher's Note:
1. Embed the tetrahedron in a cube: opposite edges are face diagonals of opposite faces, a cube-edge apart.
2. Edge L is the face diagonal, so the cube edge is L/√2.

Answer: (c) \( \frac12\overrightarrow{AC} \)

Explanation:
1. \( \overrightarrow{AD}=\frac{\vec b+\vec c}{2}-\vec a,\ \overrightarrow{BE}=\frac{\vec c+\vec a}{2}-\vec b,\ \overrightarrow{CF}=\frac{\vec a+\vec b}{2}-\vec c \).
2. Coefficient of a: \( -1+\frac13+\frac16=-\frac12 \); of b: \( \frac12-\frac23+\frac16=0 \); of c: \( \frac12+\frac13-\frac13=\frac12 \).
3. The result is \( \frac{\vec c-\vec a}{2}=\frac12\overrightarrow{AC} \).

Teacher's Note:
1. Write each median in terms of the position vectors of A, B, C.
2. Collect the coefficients of a, b and c separately.

Answer: (b) \( -\frac32 \)

Explanation:
1. Collinear: \( \frac{x+4}{2-2x}=\frac12 \) (compare the coefficients of a and of b).
2. \( 2x+8=2-2x\Rightarrow4x=-6\Rightarrow x=-\frac32 \).

Teacher's Note:
1. For non-collinear a and b, two vectors in the plane are parallel when their coefficient ratios are equal.
2. Cross-multiply carefully.

Answer: (c) \( 5\vec a \)

Explanation:
1. Let the vertices have position vectors \( \vec a=\vec v_1,\vec v_2,\dots,\vec v_5 \) about O; \( \sum\vec v_k=\vec0 \) for a regular polygon.
2. The sum is \( (\vec v_2-\vec v_1)+2(\vec v_3-\vec v_2)+3(\vec v_4-\vec v_3)+4(\vec v_5-\vec v_4)+5(\vec v_1-\vec v_5) \).
3. Collecting the coefficient of each vertex: \( \vec v_1:\ -1+5=4,\ \ \vec v_2:\ 1-2=-1,\ \ \vec v_3:\ 2-3=-1,\ \ \vec v_4:\ 3-4=-1,\ \ \vec v_5:\ 4-5=-1 \).
4. So the sum is \( 4\vec v_1-(\vec v_2+\vec v_3+\vec v_4+\vec v_5)=5\vec v_1-\sum\vec v_k=5\vec a \).

Teacher's Note:
1. Each vertex appears twice, once with its coefficient and once with the next one, so most vertices end up with coefficient −1.
2. Adding and subtracting v1 turns that into minus the vertex sum, which is zero.

Answer: (a) \( 1+\cos2A+\cos2B+\cos2C=0 \)

Explanation:
1. Let \( \hat u,\hat v \) be the unit vectors. \( \overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=\hat u+\hat v \).
2. Then \( \overrightarrow{AB}\cdot\overrightarrow{BC}=(\hat u-\hat v)\cdot(\hat u+\hat v)=1-1=0 \), so angle B = 90°.
3. Then \( C=90^\circ-A \), so \( \cos2C=-\cos2A \) and \( \cos2B=-1 \).
4. Hence \( 1+\cos2A+\cos2B+\cos2C=1+(\cos2A-\cos2A)-1=0 \).

Teacher's Note:
1. Check which angle is a right angle by computing a dot product of two sides.
2. In a right triangle cos2A + cos2C = 0.

Answer: (a) 3, −4

Explanation:
1. \( \vec b=2\vec c+\lambda\vec a \), so \( |\vec b|^2=4+\lambda^2+4\lambda(\vec a\cdot\vec c)=4+\lambda^2+\lambda \).
2. Setting it equal to 16: \( \lambda^2+\lambda-12=0\Rightarrow(\lambda+4)(\lambda-3)=0 \).
3. So \( \lambda=3\text{ or }-4 \).

Teacher's Note:
1. Square the relation and use the given dot product a·c = 1/4.
2. Both roots are valid.

Answer: (b) −25

Explanation:
1. Square the relation: \( 9+16+25+2S=0 \), so \( S=-25 \).

Teacher's Note:
1. For three vectors summing to zero, S = −(|u|² + |v|² + |w|²)/2.
2. Here the vectors form a 3-4-5 triangle, so u ⊥ v (u·v = 0) while the other two products are −16 and −9.

Answer: (a) \( \frac1{14} \)

Explanation:
1. Four points are coplanar and the combination equals zero only if the coefficients add up to zero: \( \sin\alpha+2\sin2\beta+3\sin3\gamma-1=0 \).
2. Let \( x=\sin\alpha,\ y=\sin2\beta,\ z=\sin3\gamma \). Then \( x+2y+3z=1 \).
3. By Cauchy-Schwarz \( 1=(x+2y+3z)^2\le(1+4+9)(x^2+y^2+z^2) \), so \( x^2+y^2+z^2\ge\frac1{14} \).
4. This is attained at \( (x,y,z)=\frac1{14}(1,2,3) \), which lies within [−1, 1].

Teacher's Note:
1. The condition on the coefficients (sum zero) is what makes the four points coplanar.
2. Use (x + 2y + 3z)² ≤ 14(x² + y² + z²).

Answer: (a) \( \vec r=(\vec r\cdot\hat a)\hat a+(\vec r\cdot\hat b)\hat b+(\vec r\cdot(\hat a\times\hat b))(\hat a\times\hat b) \)

Explanation:
1. \( \hat a,\hat b,\hat a\times\hat b \) are three mutually perpendicular unit vectors, so they form an orthonormal basis.
2. Any vector is the sum of its components along these: \( \vec r=(\vec r\cdot\hat a)\hat a+(\vec r\cdot\hat b)\hat b+(\vec r\cdot\hat n)\hat n \) with \( \hat n=\hat a\times\hat b \).

Teacher's Note:
1. The three components must all carry plus signs because each is the actual projection.
2. Option (d) misses the third direction and is not valid for a general vector.

Answer: (a) \( \frac1{\sqrt2}(4\hat i-\hat j-\hat k) \)

Explanation:
1. \( |OA|=3 \). The component of \( \hat i \) perpendicular to OA is \( \hat i-\frac19(\hat i+2\hat j+2\hat k)=\frac19(8,-2,-2) \), along the direction \( (4,-1,-1)/\sqrt{18} \).
2. The rotated vector has length 3: \( 3\cdot\frac{(4,-1,-1)}{3\sqrt2}=\frac1{\sqrt2}(4\hat i-\hat j-\hat k) \).

Teacher's Note:
1. The new vector is perpendicular to OA and lies in the plane of OA and i, at the same length 3.
2. Check the dot product with OA: 4 − 2 − 2 = 0.

Answer: (c) \( \lambda\hat i+(2\lambda+1)\hat j+\lambda\hat k \)

Explanation:
1. Let \( \vec n=(1,2,1) \) and \( \vec m=(1,0,-1) \). If \( \vec a\times\vec n=\vec m \), then \( \vec a=\frac{\vec n\times\vec m}{|\vec n|^2}+t\vec n \) (a particular solution plus any multiple of n).
2. Here \( \vec n\times\vec m=(-2,2,-2) \), so \( \vec a=\left(-\frac13,\frac13,-\frac13\right)+t(1,2,1) \).
3. Put \( \lambda=t-\frac13 \): \( \vec a=(\lambda,\ 2\lambda+1,\ \lambda) \).
4. Check with λ = 0: \( \hat j\times(\hat i+2\hat j+\hat k)=-\hat k+\hat i \) ✓.

Teacher's Note:
1. Parallel vectors have zero cross product, so you can add any multiple of n.
2. Verify with a simple value of the parameter.

Answer: (a) \( -\frac12 \)

Explanation:
1. \( \overrightarrow{OP}\times\overrightarrow{OQ}=(2\lambda,-2\lambda,-6) \), which must be parallel to \( (1,-1,6) \).
2. So \( \frac{2\lambda}{1}=\frac{-6}{6} \), giving \( \lambda=-\frac12 \).

Teacher's Note:
1. The normal to the plane through O, P, Q is OP × OQ.
2. Compare the third component with the first.