JEE Mathematics Vector Algebra MCQs Set 27

Answer: (d) \( r\ge5\sqrt{\frac32} \)

Explanation:
1. \( \vec a\times\vec b=(x+2)\hat i+(x-3)\hat j-5\hat k \), so \( |\vec a\times\vec b|^2=(x+2)^2+(x-3)^2+25=2x^2-2x+38=2\left(x-\frac12\right)^2+\frac{75}2 \).
2. The minimum of \( |\vec a\times\vec b| \) is \( \sqrt{\frac{75}2}=5\sqrt{\frac32} \) at \( x=\frac12 \), and it can be arbitrarily large. So \( r\ge5\sqrt{\frac32} \). Answer: (d)

Teacher's Note:
1. Compute the cross product in terms of \( x \).
2. Find the minimum of the squared magnitude by completing the square.

Answer: (d) \( \frac{2\pi}3 \)

Explanation:
1. The direction cosines satisfy \( \cos^2\frac\pi3+\cos^2\frac\pi4+\cos^2\theta=1 \).
2. \( \frac14+\frac12+\cos^2\theta=1\Rightarrow\cos\theta=\pm\frac12 \), so \( \theta=\frac\pi3 \) or \( \frac{2\pi}3 \).
3. Only \( \frac{2\pi}3 \) appears among the options. Answer: (d)

Teacher's Note:
1. Use \( l^2+m^2+n^2=1 \).
2. Solve for \( \theta \) in \( (0,\pi) \).

Answer: (d) 7

Explanation:
1. Let \( P(-1,2,6) \), \( Q(2,3,-4) \) and \( \vec n=6\hat i+3\hat j-4\hat k \). Then \( \overrightarrow{PQ}=3\hat i+\hat j-10\hat k \).
2. \( \overrightarrow{PQ}\times\vec n=26\hat i-48\hat j+3\hat k \), with magnitude squared \( 676+2304+9=2989 \), and \( |\vec n|^2=61 \).
3. Distance \( =\frac{|\overrightarrow{PQ}\times\vec n|}{|\vec n|}=\sqrt{\frac{2989}{61}}=\sqrt{49}=7 \). Answer: (d)

Teacher's Note:
1. Use distance \( =|PQ\times n|/|n| \).
2. Compute the cross product carefully.

Answer: (b) \( 4(2\hat i-2\hat j-\hat k) \)

Explanation:
1. \( \vec a+\vec b=4\hat i+4\hat j \) and \( \vec a-\vec b=2\hat i+4\hat k \), so \( (\vec a+\vec b)\times(\vec a-\vec b)=16\hat i-16\hat j-8\hat k=8(2\hat i-2\hat j-\hat k) \).
2. The unit vector along it is \( \pm\frac{2\hat i-2\hat j-\hat k}3 \), so a vector of magnitude 12 is \( \pm4(2\hat i-2\hat j-\hat k) \). Answer: (b)

Teacher's Note:
1. Find \( \vec a\pm\vec b \).
2. Take their cross product and scale to magnitude 12.

Answer: (b) is empty

Explanation:
1. \( [\vec a\ \vec b\ \vec c]=\begin{vmatrix}\alpha&1&3\\2&1&-\alpha\\\alpha&-2&3\end{vmatrix}=\alpha(3-2\alpha)-(6+\alpha^2)+3(-4-\alpha)=-3\alpha^2-18 \).
2. \( -3(\alpha^2+6)\ne0 \) for every real \( \alpha \), so the vectors are never coplanar and \( S \) is empty. Answer: (b)

Teacher's Note:
1. Set the scalar triple product equal to zero.
2. Check whether the equation has a real solution.

Answer: (a) 6

Explanation:
1. \( |\vec a|=2 \). The projection vector of \( \vec b \) on \( \vec a \) equals \( \vec a \) exactly when the scalar projection is 2: \( \frac{\vec b\cdot\vec a}{|\vec a|}=\frac{b_1+b_2+2}2=2 \), so \( b_1+b_2=2 \).
2. \( (\vec a+\vec b)\cdot\vec c=0 \): \( 5(b_1+1)+(b_2+1)+2\sqrt2\cdot\sqrt2=0\Rightarrow5b_1+b_2=-10 \).
3. Solving: \( b_1=-3 \), \( b_2=5 \), so \( |\vec b|^2=9+25+2=36 \) and \( |\vec b|=6 \). Answer: (a)

Teacher's Note:
1. Use the scalar projection of \( \vec b \) on \( \vec a \).
2. Use the perpendicularity condition.

Answer: (c) \( \left(-\frac12,4,0\right) \)

Explanation:
1. \( \vec b=2\vec a \) gives \( 3-\lambda_2=2\lambda_1 \), i.e. \( 2\lambda_1+\lambda_2=3 \).
2. \( \vec a\cdot\vec c=0 \): \( 6+6\lambda_1+3(\lambda_3-1)=0\Rightarrow2\lambda_1+\lambda_3=-1 \).
3. Test the options: only \( \left(-\frac12,4,0\right) \) satisfies both (\( -1+4=3 \) and \( -1+0=-1 \)). Answer: (c)

Teacher's Note:
1. Compare components in \( \vec b=2\vec a \).
2. Use \( \vec a\cdot\vec c=0 \).

Answer: (a) \( -10\hat i+5\hat j \)

Explanation:
1. Coplanarity: \( \begin{vmatrix}1&2&4\\1&\lambda&4\\2&4&\lambda^2-1\end{vmatrix}=(\lambda-2)(\lambda^2-9)=0 \), so \( \lambda=2,3,-3 \).
2. For \( \lambda=\pm3 \), \( \vec c=2\vec a \), so \( \vec a\times\vec c=\vec 0 \). For \( \lambda=2 \), \( \vec c=2\hat i+4\hat j+3\hat k \) and \( \vec a\times\vec c=-10\hat i+5\hat j \). Answer: (a)

Teacher's Note:
1. Use the triple product to find \( \lambda \).
2. Reject the values that give the zero vector.

Answer: (d) -1

Explanation:
1. Coplanar: \( \begin{vmatrix}\mu&1&1\\1&\mu&1\\1&1&\mu\end{vmatrix}=\mu^3-3\mu+2=(\mu-1)^2(\mu+2)=0 \).
2. The distinct values are \( \mu=1 \) and \( \mu=-2 \), with sum \( -1 \). Answer: (d)

Teacher's Note:
1. Set the determinant to zero.
2. Factorise the cubic.

Answer: (a) \( 30^\circ \)

Explanation:
1. \( \vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c=\frac12\vec b \). Since \( \vec b,\vec c \) are independent, \( \vec a\cdot\vec c=\frac12 \) and \( \vec a\cdot\vec b=0 \).
2. So \( \cos\beta=\frac12\Rightarrow\beta=\frac\pi3 \) and \( \cos\alpha=0\Rightarrow\alpha=\frac\pi2 \).
3. \( |\alpha-\beta|=\frac\pi6=30^\circ \). Answer: (a)

Teacher's Note:
1. Expand the vector triple product.
2. Compare coefficients of the independent vectors \( \vec b,\vec c \).

Answer: (b) 2

Explanation:
1. \( \vec a\times\vec b=2\hat i-2\hat j+\hat k \), so \( |\vec a\times\vec b|=3 \). Also \( |\vec a|=3 \).
2. \( |(\vec a\times\vec b)\times\vec c|=|\vec a\times\vec b||\vec c|\sin30^\circ=3\Rightarrow|\vec c|=2 \).
3. \( |\vec c-\vec a|^2=|\vec c|^2+|\vec a|^2-2\vec a\cdot\vec c=9\Rightarrow4+9-2\vec a\cdot\vec c=9 \), so \( \vec a\cdot\vec c=2 \). Answer: (b)

Teacher's Note:
1. Find \( |\vec a\times\vec b| \).
2. Get \( |\vec c| \) from the cross product magnitude.

Answer: (a) \( \frac{2\sqrt2}3 \)

Explanation:
1. \( (\vec a\times\vec b)\times\vec c=(\vec a\cdot\vec c)\vec b-(\vec b\cdot\vec c)\vec a=\frac13|\vec b||\vec c|\vec a \).
2. Since \( \vec a,\vec b \) are not collinear: \( \vec a\cdot\vec c=0 \) and \( -\vec b\cdot\vec c=\frac13|\vec b||\vec c| \), so \( \cos\theta=-\frac13 \).
3. \( \sin\theta=\sqrt{1-\frac19}=\frac{2\sqrt2}3 \) (taking \( \theta\in(0,\pi) \)). Answer: (a)

Teacher's Note:
1. Expand the triple product.
2. Compare coefficients of the independent vectors \( \vec a,\vec b \).

Answer: (b) 1

Explanation:
1. Use the identity \( [\vec a\times\vec b\ \ \vec b\times\vec c\ \ \vec c\times\vec a]=[\vec a\ \vec b\ \vec c]^2 \).
2. Comparing with \( \lambda[\vec a\ \vec b\ \vec c]^2 \) gives \( \lambda=1 \). Answer: (b)

Teacher's Note:
1. Use the triple product of cross products identity.
2. Compare both sides.

Answer: (c) \( \frac\pi3 \)

Explanation:
1. \( \vec c\cdot\vec d=0\Rightarrow(\vec a+2\vec b)\cdot(5\vec a-4\vec b)=5-4\vec a\cdot\vec b+10\vec a\cdot\vec b-8=0 \).
2. \( 6\vec a\cdot\vec b=3\Rightarrow\cos\theta=\frac12 \), so \( \theta=\frac\pi3 \). Answer: (c)

Teacher's Note:
1. Expand the dot product using \( |\vec a|=|\vec b|=1 \).
2. Solve for \( \vec a\cdot\vec b \).

Answer: (d) -5

Explanation:
1. \( |\vec a|=|\vec b|=1 \) and \( \vec a\cdot\vec b=\frac{6-6}{7\sqrt{10}}=0 \).
2. \( (\vec a\times\vec b)\times\vec a=\vec b \) and \( (\vec a\times\vec b)\times\vec b=-\vec a \), so \( (\vec a\times\vec b)\times(\vec a+2\vec b)=\vec b-2\vec a \).
3. The value is \( (2\vec a-\vec b)\cdot(\vec b-2\vec a)=-|2\vec a-\vec b|^2=-(4+1-0)=-5 \). Answer: (d)

Teacher's Note:
1. Check that \( \vec a,\vec b \) are orthogonal unit vectors.
2. Expand the triple product using \( (\vec x\times\vec y)\times\vec z \).

Answer: (c) \( \vec c-\left(\frac{\vec a\cdot\vec c}{\vec a\cdot\vec b}\right)\vec b \)

Explanation:
1. \( \vec b\times(\vec c-\vec d)=\vec 0 \), so \( \vec c-\vec d=\lambda\vec b \) and \( \vec d=\vec c-\lambda\vec b \).
2. \( \vec a\cdot\vec d=0\Rightarrow\vec a\cdot\vec c-\lambda\,\vec a\cdot\vec b=0\Rightarrow\lambda=\frac{\vec a\cdot\vec c}{\vec a\cdot\vec b} \).
3. \( \vec d=\vec c-\frac{\vec a\cdot\vec c}{\vec a\cdot\vec b}\vec b \). Answer: (c)

Teacher's Note:
1. Use the fact that the cross product of parallel vectors is zero.
2. Impose \( \vec a\cdot\vec d=0 \).

Answer: (d) \( \vec 0 \)

Explanation:
1. \( \vec a+3\vec b=\lambda\vec c \) and \( \vec b+2\vec c=\mu\vec a \). From the second, \( \vec b=\mu\vec a-2\vec c \).
2. Substitute: \( \vec a+3\mu\vec a-6\vec c=\lambda\vec c \), so \( (1+3\mu)\vec a=(\lambda+6)\vec c \). Since \( \vec a,\vec c \) are not collinear, both coefficients vanish: \( \lambda=-6 \).
3. Hence \( \vec a+3\vec b=-6\vec c \), i.e. \( \vec a+3\vec b+6\vec c=\vec 0 \). Answer: (d)

Teacher's Note:
1. Write both collinearity conditions with scalars.
2. Use independence of non-collinear vectors.

Answer: (a) (-3, 2)

Explanation:
1. \( \vec a\cdot\vec b=2-4+2=0 \), as required.
2. \( \vec a\cdot\vec c=\lambda-1+2\mu=0 \) and \( \vec b\cdot\vec c=2\lambda+4+\mu=0 \).
3. Solving: \( \mu=2 \) and \( \lambda=-3 \). Answer: (a)

Teacher's Note:
1. Set the pairwise dot products to zero.
2. Solve the two linear equations.