JEE Mathematics Vector Algebra MCQs Set 11

Answer: (d) \( \left[\dfrac{\pi}{4},\dfrac{3\pi}{4}\right] \)

Explanation:
1. \( \vec c\cdot\vec a=\vec c\cdot\vec b=x=\cos\theta \).
2. \( |\vec c|=1 \) gives \( 2x^2+y^2=1 \), so \( x^2\le\dfrac12 \), i.e. \( |\cos\theta|\le\dfrac{1}{\sqrt2} \).
3. Hence \( \theta\in\left[\dfrac{\pi}{4},\dfrac{3\pi}{4}\right] \).

Teacher's Note:
1. Take dot products with \( \vec a \) and \( \vec b \) to identify \( x \) as \( \cos\theta \).
2. Use the unit magnitude of \( \vec c \) to bound \( x \).

Answer: (c) \( 0 \)

Explanation:
1. Let \( \vec p=\vec a\times(\vec b+\vec c) \). Then \( \vec p\times(\vec b\times\vec c)=\vec b(\vec p\cdot\vec c)-\vec c(\vec p\cdot\vec b) \).
2. \( \vec p\cdot\vec c=[\vec a\ \vec b\ \vec c] \) and \( \vec p\cdot\vec b=-[\vec a\ \vec b\ \vec c] \), so \( \vec p\times(\vec b\times\vec c)=[\vec a\ \vec b\ \vec c](\vec b+\vec c) \).
3. Dotting with \( \vec b-\vec c \): \( [\vec a\ \vec b\ \vec c](|\vec b|^2-|\vec c|^2)=0 \) for unit vectors.

Teacher's Note:
1. Use \( (\vec p\times\vec q)\times\vec s \) expansions step by step.
2. The factor \( (\vec b+\vec c)\cdot(\vec b-\vec c)=|\vec b|^2-|\vec c|^2 \) vanishes for unit vectors.

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

Explanation:
1. \( [\vec a\ \vec b\ \hat i]=(\vec a\times\vec b)\cdot\hat i \), and similarly for \( \hat j,\hat k \).
2. These are the components of \( \vec a\times\vec b \) along the axes.
3. The sum is therefore \( \vec a\times\vec b \).

Teacher's Note:
1. A scalar triple product with a unit axis vector picks one component of the cross product.
2. Adding the three components rebuilds the vector.

Answer: (d) the geometric mean of \( a \) and \( b \)

Explanation:
1. Coplanarity: \( \begin{vmatrix}a&a&c\\1&0&1\\c&c&b\end{vmatrix}=0 \).
2. Expanding along the first row: \( a(0-c)-a(b-c)+c(c-0)=c^2-ab \).
3. So \( c^2=ab \) and \( c \) is the geometric mean of \( a \) and \( b \).

Teacher's Note:
1. Three vectors lie in a plane exactly when their determinant is zero.
2. Expand along the row with a zero to keep the algebra short.

Answer: (c) \( \vec v\cdot(\vec u\times\vec w) \)

Explanation:
1. \( \vec u\cdot(\vec v\times\vec w)=[\vec u\ \vec v\ \vec w] \).
2. \( (\vec v\times\vec w)\cdot\vec u \) and \( (\vec u\times\vec v)\cdot\vec w \) are cyclic arrangements, also equal to \( [\vec u\ \vec v\ \vec w] \).
3. \( \vec v\cdot(\vec u\times\vec w)=[\vec v\ \vec u\ \vec w]=-[\vec u\ \vec v\ \vec w] \).

Teacher's Note:
1. Cyclic order keeps the sign of a scalar triple product; swapping two vectors changes it.
2. Compare each option with \( [\vec u\ \vec v\ \vec w] \).

Answer: (d) \( \dfrac32 \)

Explanation:
1. \( |\vec a|=3 \). From \( |\vec c-\vec a|^2=8 \): \( |\vec c|^2-2\vec a\cdot\vec c+9=8 \), and \( \vec a\cdot\vec c=|\vec c| \) so \( (|\vec c|-1)^2=0 \), \( |\vec c|=1 \).
2. \( \vec a\times\vec b=(2,-2,1) \), magnitude 3.
3. \( |(\vec a\times\vec b)\times\vec c|=3\cdot1\cdot\sin30^\circ=\dfrac32 \).

Teacher's Note:
1. Convert the two conditions on \( \vec c \) into a perfect square to get its magnitude.
2. The magnitude of a cross product uses the sine of the angle.

Answer: (c) \( \dfrac{1}{\sqrt2}(-\hat j+\hat k) \)

Explanation:
1. A vector in the plane perpendicular to \( \vec a \) is \( \vec b-\dfrac{\vec a\cdot\vec b}{|\vec a|^2}\vec a=\vec b-\tfrac12\vec a=\left(0,\tfrac32,-\tfrac32\right) \).
2. Normalising gives \( \dfrac{1}{\sqrt2}(\hat j-\hat k) \), and its negative \( \dfrac{1}{\sqrt2}(-\hat j+\hat k) \) is also valid.

Teacher's Note:
1. Remove from \( \vec b \) its component along \( \vec a \) to get the perpendicular part in the plane.
2. Remember a unit vector can have either sign.

Answer: (c) \( \sqrt{59} \)

Explanation:
1. \( [\vec U\ \vec V\ \vec W]=\vec U\cdot(\vec V\times\vec W) \), maximum when \( \vec U \) is along \( \vec V\times\vec W \).
2. \( \vec V\times\vec W=(3,-7,-1) \), magnitude \( \sqrt{9+49+1}=\sqrt{59} \).

Teacher's Note:
1. The dot product with a unit vector is largest when the vectors are parallel.
2. So the maximum equals the magnitude of the cross product.

Answer: (b) \( \dfrac{1}{\sqrt3} \)

Explanation:
1. Volume \( =\begin{vmatrix}1&a&1\\0&1&a\\a&0&1\end{vmatrix}=1+a^3-a \) (taken positive for \( a>0 \)).
2. \( \dfrac{d}{da}(a^3-a+1)=3a^2-1=0 \) gives \( a=\dfrac1{\sqrt3} \), and the second derivative \( 6a>0 \) confirms a minimum.

Teacher's Note:
1. Write the volume as a determinant, then differentiate.
2. Check the sign of the second derivative to confirm a minimum.

Answer: (d) \( \dfrac{3\hat j-\hat k}{\sqrt{10}} \)

Explanation:
1. A coplanar vector is \( \lambda(2,1,1)+\mu(1,-1,1) \). Perpendicular to \( (3,2,6) \): \( 3(2\lambda+\mu)+2(\lambda-\mu)+6(\lambda+\mu)=14\lambda+7\mu=0 \), so \( \mu=-2\lambda \).
2. The vector is \( \lambda(0,3,-1) \); normalising gives \( \dfrac{3\hat j-\hat k}{\sqrt{10}} \).

Teacher's Note:
1. Write the general vector in the plane with two parameters.
2. Impose the perpendicularity condition to relate them.

Answer: (a) \( 4\hat i-\hat j+4\hat k \)

Explanation:
1. For \( 4\hat i-\hat j+4\hat k \): \( \det[\vec a,\vec b,\cdot]=0 \) so it is coplanar, and \( (4,-1,4)\cdot(1,1,-1)=-1 \), giving projection magnitude \( \dfrac1{\sqrt3} \).
2. For \( 4\hat i+\hat j-4\hat k \) the determinant is 24 (not coplanar) and for \( 2\hat i+\hat j+\hat k \) it is 3 (not coplanar).

Teacher's Note:
1. Test each option for both conditions: coplanarity by a determinant, projection by a dot product.
2. Any option failing one test is eliminated.

Answer: (c) \( 2 \)

Explanation:
1. Let \( t=\lambda^2 \). The determinant \( \begin{vmatrix}-t&1&1\\1&-t&1\\1&1&-t\end{vmatrix}=-(t+1)^2(t-2) \) must vanish.
2. \( t=-1 \) is impossible for real \( \lambda \), so \( t=2 \), i.e. \( \lambda=\pm\sqrt2 \).
3. Two distinct values.

Teacher's Note:
1. Substitute \( t=\lambda^2 \) to make the determinant easy to factor.
2. Reject roots that give a negative square.

Answer: (a) \( \dfrac{\beta\vec a-\vec a\times\vec c}{|\vec a|^2} \)

Explanation:
1. \( \vec a\times\vec c=\vec a\times(\vec a\times\vec b)=\vec a(\vec a\cdot\vec b)-\vec b|\vec a|^2=\beta\vec a-|\vec a|^2\vec b \).
2. Hence \( \vec b=\dfrac{\beta\vec a-\vec a\times\vec c}{|\vec a|^2} \).

Teacher's Note:
1. Cross the given relation with \( \vec a \) to eliminate the cross product.
2. Use the triple product expansion \( \vec a\times(\vec a\times\vec b) \).

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

Explanation:
1. \( \vec d\times(\vec a\times\vec d)=\vec a|\vec d|^2-\vec d(\vec d\cdot\vec a) \), so the expression is \( \vec a-\dfrac{(\vec a\cdot\vec d)}{|\vec d|^2}\vec d \), the component of \( \vec a \) perpendicular to \( \vec d \).
2. \( \vec b\parallel\vec d \) and \( \vec c\perp\vec d \), so the perpendicular component of \( \vec a=\vec b+\vec c \) is \( \vec c \).

Teacher's Note:
1. The expression is the part of \( \vec a \) perpendicular to \( \vec d \).
2. Split \( \vec a \) into the part parallel to \( \vec d \) and the part perpendicular to it.

Answer: (b) \( \dfrac{3\pi}{4} \)

Explanation:
1. \( \vec a\times(\vec b\times\vec c)=\vec b(\vec a\cdot\vec c)-\vec c(\vec a\cdot\vec b) \).
2. As \( \vec b,\vec c \) are independent, \( \vec a\cdot\vec c=\dfrac1{\sqrt2} \) and \( \vec a\cdot\vec b=-\dfrac1{\sqrt2} \).
3. \( \cos\theta=-\dfrac1{\sqrt2} \), so the angle between \( \vec a \) and \( \vec b \) is \( \dfrac{3\pi}{4} \).

Teacher's Note:
1. Expand the triple cross product and compare coefficients of \( \vec b \) and \( \vec c \).
2. Independence of \( \vec b \) and \( \vec c \) is what allows the comparison.

Answer: (c) \( \tfrac12(\vec p+\vec q+\vec r) \)

Explanation:
1. Take \( |\vec p|=|\vec q|=|\vec r|=1 \) (the equation is homogeneous). Then \( \vec p\times(\vec y\times\vec p)=\vec y-\vec p(\vec p\cdot\vec y) \), and \( \vec p\cdot(\vec x-\vec q)=\vec p\cdot\vec x \).
2. The sum is \( 3\vec x-(\vec p+\vec q+\vec r)-[\vec p(\vec p\cdot\vec x)+\vec q(\vec q\cdot\vec x)+\vec r(\vec r\cdot\vec x)]=3\vec x-(\vec p+\vec q+\vec r)-\vec x \).
3. Setting it to zero gives \( \vec x=\tfrac12(\vec p+\vec q+\vec r) \).

Teacher's Note:
1. For an orthonormal triple, the sum of projections reproduces the vector itself.
2. Expand each triple cross product with the standard identity.

Answer: (b) \( 0 \)

Explanation:
1. The normals of the planes are \( \vec a\times\vec b \) and \( \vec c\times\vec d \).
2. Their cross product is zero, so the normals are parallel.
3. Parallel normals mean parallel planes, so the angle is 0.

Teacher's Note:
1. The angle between planes is the angle between their normals.
2. A zero cross product means parallel vectors.

Answer: (b) (i) and (iii) only

Explanation:
1. With the vertices as \( A,B,C \): \( \overrightarrow{AB}=(-1,-1,-4) \), \( \overrightarrow{BC}=(-3,3,0) \), \( \overrightarrow{AC}=(-4,2,-4) \).
2. \( AB^2=18,\ BC^2=18,\ AC^2=36 \).
3. \( AB=BC \) so it is isosceles, and \( AB^2+BC^2=AC^2 \) so it is right-angled at B. Answer: (i) and (iii).

Teacher's Note:
1. Compute all three side lengths before classifying.
2. Check the Pythagoras relation for a right angle.

Answer: (a) \( 0 \) and \( -1 \)

Explanation:
1. The equations are \( x+3y-4z=\lambda x \), \( x-3y+5z=\lambda y \), \( 3x+y=\lambda z \).
2. This is an eigenvalue problem for \( \begin{pmatrix}1&3&-4\\1&-3&5\\3&1&0\end{pmatrix} \), and its eigenvalues (checked with sympy) are \( -1 \) (twice) and \( 0 \).
3. So the possible values are \( 0 \) and \( -1 \). Answer: 0 and -1.

Teacher's Note:
1. Treat the system as \( (A-\lambda I)\vec v=0 \) and use the determinant.
2. Only eigenvalues give non-zero solutions.

Answer: (d) (i) only

Explanation:
1. Coplanarity needs \( \begin{vmatrix}1&2&3\\0&\lambda&\mu\\0&0&2\lambda-1\end{vmatrix}=\lambda(2\lambda-1)=0 \).
2. So \( \lambda=0 \) or \( \lambda=\tfrac12 \), for every \( \mu \). Answer: (i) only.

Teacher's Note:
1. The coefficient determinant is triangular here.
2. Its zeros give the values of \( \lambda \).

Answer: (d) \( 2 \) and \( -\dfrac23 \)

Explanation:
1. Doubling the magnitude means \( 16+(4x-2)^2+4=4(1+x^2+9) \).
2. \( 16x^2-16x+24=4x^2+40 \), so \( 3x^2-4x-4=0 \).
3. \( x=2 \) or \( x=-\dfrac23 \).

Teacher's Note:
1. Rotation keeps direction changes free, so only the magnitude condition is needed.
2. Solve the resulting quadratic.

Answer: (c) (i) and (iv)

Explanation:
1. Squaring: \( |\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2=|\vec a|^2-4\vec a\cdot\vec b+4|\vec b|^2 \), so \( 2\vec a\cdot\vec b=|\vec b|^2 \).
2. Let \( t=|\vec b|^2 \). Then \( \vec a\cdot\vec b+\dfrac1{t+2}=\dfrac t2+\dfrac1{t+2}=\dfrac{t+2}2+\dfrac1{t+2}-1 \).
3. By AM-GM this is at least \( 2\sqrt{\tfrac12}-1=\sqrt2-1 \). Answer: (i) and (iv).

Teacher's Note:
1. Square both sides and cancel the common term.
2. Rewrite the expression so that AM-GM applies.