JEE Mathematics Vector Algebra MCQs Set 13

Answer: (b) \( 3 \)

Explanation:
1. \( (\vec a\times\vec b)\cdot(\vec c\times\vec d)=(\vec a\cdot\vec c)(\vec b\cdot\vec d)-(\vec a\cdot\vec d)(\vec b\cdot\vec c) \).
2. \( \vec a\cdot\vec c=4 \), \( \vec b\cdot\vec d=-3 \), \( \vec a\cdot\vec d=11 \), \( \vec b\cdot\vec c=-3 \).
3. \( 4(-3)-11(-3)=21 \), and \( \dfrac{21}7=3 \).

Teacher's Note:
1. Use the Lagrange identity for the dot product of two cross products.
2. Evaluate the four dot products separately.

Answer: (c) \( \sqrt{57} \)

Explanation:
1. \( |\vec a-\vec b|^2+|\vec a+\vec b|^2=2(|\vec a|^2+|\vec b|^2) \).
2. \( 64+100=2(25+|\vec b|^2) \), so \( |\vec b|^2=57 \).
3. \( |\vec b|=\sqrt{57} \).

Teacher's Note:
1. Add the expansions of \( |\vec a\pm\vec b|^2 \) to cancel the dot product.
2. This is the parallelogram law.

Answer: (b) \( 24 \)

Explanation:
1. \( [\vec a\times\vec b,\ \vec b\times\vec c,\ \vec c\times\vec a]=[\vec a\ \vec b\ \vec c]^2=4 \).
2. With the scalar factors \( 2\cdot3\cdot1=6 \): volume \( =6\cdot4=24 \).

Teacher's Note:
1. The triple product of the three cross products equals the square of \( [\vec a\ \vec b\ \vec c] \).
2. Multiply by the scalar factors in the vectors.

Answer: (d) \( 60 \)

Explanation:
1. \( [\vec a+\vec b,\ \vec b+\vec c,\ \vec c+\vec a]=2[\vec a\ \vec b\ \vec c] \).
2. With the factors \( 3\cdot1\cdot2=6 \): volume \( =6\cdot2\cdot5=60 \).

Teacher's Note:
1. The determinant of coefficients \( \begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}=2 \) multiplies the triple product.
2. Scale factors multiply as well.

Answer: (c) \( 100 \)

Explanation:
1. \( \tfrac12|\vec a\times\vec b|=20 \), so \( |\vec a\times\vec b|=40 \).
2. \( (2\vec a+3\vec b)\times(\vec a-\vec b)=-2\,\vec a\times\vec b+3\,\vec b\times\vec a=-5\,\vec a\times\vec b \).
3. Area \( =\tfrac12\cdot5\cdot40=100 \).

Teacher's Note:
1. Expand the cross product and use \( \vec b\times\vec a=-\vec a\times\vec b \).
2. Area of a triangle is half the magnitude of the cross product.

Answer: (d) \( -\dfrac32 \)

Explanation:
1. \( |\vec a+\vec b+\vec c|^2=0=3+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a) \).
2. Hence the sum of dot products is \( -\dfrac32 \).

Teacher's Note:
1. Square the zero vector relation.
2. The dot products appear with a factor 2.

Answer: (b) \( c\in\left(-\dfrac43,0\right) \)

Explanation:
1. Let \( L=\log_2x \), which takes every real value. \( \vec p\cdot\vec q=cL^2-12+6cL<0 \) for all \( L \).
2. This needs \( c<0 \) and discriminant \( 36c^2+48c<0 \), i.e. \( -\dfrac43<c<0 \).

Teacher's Note:
1. A quadratic in \( L \) is negative for all \( L \) only if its leading coefficient is negative and its discriminant is negative.
2. Here \( x\in(0,\infty) \) makes \( L \) range over all reals.

Answer: (a) \( 3 \)

Explanation:
1. With A as origin, \( \overrightarrow{AA_1}=\vec b+\tfrac12\vec d \) and \( \overrightarrow{AB_1}=\vec d+\tfrac12\vec b \), where \( \vec b=\overrightarrow{AB},\vec d=\overrightarrow{AD} \).
2. The sum is \( \tfrac32(\vec b+\vec d)=\tfrac32\overrightarrow{AC} \), so \( \lambda=\tfrac32 \).
3. \( 2\lambda=3 \).

Teacher's Note:
1. Express all vectors through two adjacent sides.
2. The diagonal AC is the sum of those sides.

Answer: (b) \( 6 \)

Explanation:
1. For any point P: \( PA^2+PB^2+PC^2=3PG^2+GA^2+GB^2+GC^2 \), and \( GA^2+GB^2+GC^2=\dfrac{a^2+b^2+c^2}3 \).
2. \( 1+9+4=3\cdot4+\dfrac{a^2+b^2+c^2}3 \) gives \( 14=12+\dfrac{a^2+b^2+c^2}3 \).
3. So \( a^2+b^2+c^2=6 \).

Teacher's Note:
1. Use the identity connecting distances from a point to the vertices and to the centroid.
2. The sum of squared medians-to-centroid distances is one third of the sum of squared sides.

Answer: (b) \( 5 \)

Explanation:
1. \( \vec p\cdot\vec q=\dfrac{2\sqrt2}3\cdot\dfrac1{\sqrt2}=\dfrac23 \) and \( |\vec p|^2=\dfrac89 \).
2. \( |\vec a+\vec b|^2=|6\vec p-\vec q|^2=32+1-12\cdot\dfrac23=25 \).
3. \( |\vec b-\vec a|^2=|4\vec p+5\vec q|^2=\dfrac{128}9+25+\dfrac{80}3=\dfrac{593}9\approx65.9 \).
4. The shorter diagonal is 5.

Teacher's Note:
1. Diagonals are the sum and difference of the adjacent sides.
2. Compare the two lengths numerically.

Answer: (a) \( 3 \)

Explanation:
1. \( \vec a\cdot\vec b=x(x+1)+(x-1)+a=x^2+2x-1+a>0 \) for all \( x \).
2. The discriminant must be negative: \( 4-4(a-1)<0 \), so \( a>2 \).
3. The least integer is 3.

Teacher's Note:
1. A quadratic is positive for all \( x \) when its discriminant is negative and the leading coefficient is positive.
2. Note that \( a=2 \) gives a zero dot product at \( x=-1 \), so it is excluded.

Answer: (c) \( 3 \)

Explanation:
1. The left side \( =6-2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=9 \), so the sum of dot products is \( -\dfrac32 \).
2. \( |\vec a+\vec b+\vec c|^2=3+2\left(-\dfrac32\right)=0 \), so \( \vec b+\vec c=-\vec a \).
3. \( 2\vec a+5(\vec b+\vec c)=-3\vec a \), with magnitude 3.

Teacher's Note:
1. Expand the sum of squared differences using unit magnitudes.
2. A zero magnitude of the sum gives a linear relation.

Answer: (b) \( 2 \)

Explanation:
1. \( |\vec a+\vec b|^2=3 \) gives \( 2+2\vec a\cdot\vec b=3 \), so \( \vec a\cdot\vec b=\tfrac12 \).
2. \( \vec c\cdot\vec b=\vec a\cdot\vec b+2|\vec b|^2+3(\vec a\times\vec b)\cdot\vec b=\tfrac12+2+0=\tfrac52 \).
3. The greatest integer of 2.5 is 2.

Teacher's Note:
1. A cross product is perpendicular to both of its factors, so it drops out of the dot product with \( \vec b \).
2. Apply the floor function at the end.

Answer: (c) \( 3 \)

Explanation:
1. The weights 1, 2, 3 at A, B, C mean the areas of triangles OBC, OCA, OAB are proportional to 1, 2, 3.
2. Total \( =6 \) and triangle AOC has weight 2, so the ratio is \( \dfrac62=3 \).

Teacher's Note:
1. For \( \alpha\vec{OA}+\beta\vec{OB}+\gamma\vec{OC}=0 \), the area of triangle opposite a vertex is proportional to that vertex's weight.
2. Triangle AOC is the one opposite B, with weight 2.

Answer: (a) \( 8 \)

Explanation:
1. With circumradius \( R \) and consecutive vectors at angle \( \dfrac{2\pi}n \): the left side is \( (n-1)R^2\sin\dfrac{2\pi}n \) and the right side is \( (n-1)R^2\left|\cos\dfrac{2\pi}n\right| \).
2. So \( \left|\tan\dfrac{2\pi}n\right|=1 \), giving \( \dfrac{2\pi}n=\dfrac\pi4 \) or \( \dfrac{3\pi}4 \) (for \( n>2 \)).
3. \( n=8 \) or \( n=\dfrac83 \), and the only integer is \( n=8 \), the maximum.

Teacher's Note:
1. Each cross product has magnitude \( R^2\sin\dfrac{2\pi}n \) and each dot product \( R^2\cos\dfrac{2\pi}n \).
2. Set the tangent equal to 1 and test integer values of n.

Answer: (b) \( 2 \)

Explanation:
1. Coplanar: \( \det[B-A,\ C-A,\ D-A]=0 \) with \( B-A=(-1,5,-3) \), \( C-A=(-4,3,3) \), \( D-A=(1,7,\lambda+1) \).
2. Solving (sympy-checked) gives \( \lambda=-\dfrac{146}{17} \).
3. \( \dfrac{-17\lambda}{73}=\dfrac{146}{73}=2 \).

Teacher's Note:
1. Four points are coplanar when the triple product of the three joining vectors is zero.
2. Solve the linear equation for the unknown component.

Answer: (d) \( 3 \)

Explanation:
1. The first term is \( 2[\vec a\ \vec b\ \vec c]=2 \). The second is \( [\vec a\ \vec b\ \vec c]^2=1 \).
2. The three vectors in the third bracket add up to \( \vec 0 \) (Jacobi identity), so they are coplanar and the triple product is 0.
3. Total \( =2+1+0=3 \).

Teacher's Note:
1. Learn the standard results \( [\vec a+\vec b,\vec b+\vec c,\vec c+\vec a]=2[\vec a\,\vec b\,\vec c] \) and \( [\vec a\times\vec b,\vec b\times\vec c,\vec c\times\vec a]=[\vec a\,\vec b\,\vec c]^2 \).
2. The sum \( \vec a\times(\vec b\times\vec c)+\vec b\times(\vec c\times\vec a)+\vec c\times(\vec a\times\vec b) \) is always zero.

Answer: (a) \( 2 \)

Explanation:
1. The normals are \( (1,1,0)\times(0,1,1)=(1,-1,1) \), \( (0,1,1)\times(1,0,1)=(1,1,-1) \) and \( (1,0,1)\times(1,1,0)=(-1,1,1) \), each of length \( \sqrt3 \).
2. The determinant of the three normals is 4, so for unit vectors the triple product is \( \dfrac{4}{3\sqrt3} \).
3. \( V=\dfrac16\cdot\dfrac4{3\sqrt3}=\dfrac2{9\sqrt3} \), so \( 9\sqrt3V=2 \).

Teacher's Note:
1. The normal to a plane through two vectors is their cross product.
2. Divide each normal by its length before using the triple product.

Answer: (c) \( 6 \)

Explanation:
1. Dot the first relation with \( \vec a \): \( \ell[\vec a\ \vec b\ \vec c]=4\,a_x \), so \( \ell=2a_x \). Similarly \( m=2b_x \) and \( n=2c_x \).
2. \( \ell+m+n=2(a_x+b_x+c_x)=2\cdot3=6 \).

Teacher's Note:
1. Dotting with one of the original vectors isolates one coefficient.
2. The sum of x-components comes from \( \vec a+\vec b+\vec c=3\hat i \).

Answer: (c) \( 8 \)

Explanation:
1. \( \vec a\times\vec b=(-5,3,2) \), \( |\vec a\times\vec b|^2=38 \).
2. \( [\vec a\ \vec b\ \vec c]=(-5)(1)+3(1)+2(3)=4 \) in absolute value.
3. \( h=\dfrac{4}{\sqrt{38}} \), so \( 19h^2=\dfrac{19\cdot16}{38}=8 \).

Teacher's Note:
1. Height equals volume divided by base area.
2. Square before multiplying to avoid the root.

Answer: (d) \( 7 \)

Explanation:
1. The conditions give \( \vec a\cdot\vec c+\vec b\cdot\vec c=0 \), \( \vec a\cdot\vec b+\vec a\cdot\vec c=0 \), \( \vec a\cdot\vec b+\vec b\cdot\vec c=0 \). Adding and subtracting shows all three dot products are 0.
2. \( |\vec a+\vec b+\vec c|^2=9+16+25=50 \), so \( |\vec a+\vec b+\vec c|=\sqrt{50}\approx7.07 \).
3. The greatest integer is 7.

Teacher's Note:
1. Solve the three linear equations in the dot products.
2. Mutually perpendicular vectors add in squares of lengths.