JEE Mathematics Vector Algebra MCQs Set 29

Answer: (a) \(\frac\pi4\)

Explanation:
1. \(\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{\sqrt6}{2\sqrt3}=\frac{\sqrt2}{2}=\frac1{\sqrt2}\).
2. So \(\theta=\frac\pi4\).

Teacher's Note:
1. Use \(\cos heta=rac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\) and simplify the surd.

Answer: (b) \(|\mathbf{a}|=|\mathbf{b}|=1\)

Explanation:
1. \(\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos60^\circ=|\mathbf{a}|^2\cdot\frac12\) since the magnitudes are equal.
2. Setting this equal to \(\frac12\) gives \(|\mathbf{a}|^2=1\), so \(|\mathbf{a}|=|\mathbf{b}|=1\).

Teacher's Note:
1. Equal magnitudes reduce the scalar product to \(|\mathbf{a}|^2\cos heta\).

Answer: (a) \(\frac{16}{3}\sqrt{\frac27},\ \frac23\sqrt{\frac27}\)

Explanation:
1. \((\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}-\mathbf{b})=|\mathbf{a}|^2-|\mathbf{b}|^2=64|\mathbf{b}|^2-|\mathbf{b}|^2=63|\mathbf{b}|^2=8\).
2. \(|\mathbf{b}|^2=\frac{8}{63}\), so \(|\mathbf{b}|=\frac23\sqrt{\frac27}\) (since \(\frac{8}{63}=\frac{4}{9}\cdot\frac27\)).
3. Then \(|\mathbf{a}|=8|\mathbf{b}|=\frac{16}{3}\sqrt{\frac27}\).

Teacher's Note:
1. Expand the dot product of a sum and difference as a difference of squares.

Answer: (c) \(\sqrt{13}\)

Explanation:
1. \((\mathbf{x}-\mathbf{a})\cdot(\mathbf{x}+\mathbf{a})=|\mathbf{x}|^2-|\mathbf{a}|^2=|\mathbf{x}|^2-1=12\).
2. So \(|\mathbf{x}|^2=13\) and \(|\mathbf{x}|=\sqrt{13}\).

Teacher's Note:
1. Use \(|\mathbf{a}|=1\) for a unit vector.

Answer: (d) 8

Explanation:
1. \(\mathbf{a}+\lambda\mathbf{b}=(2-\lambda)\hat i+(2+2\lambda)\hat j+(3+\lambda)\hat k\).
2. Perpendicular to \(\mathbf{c}\): \(3(2-\lambda)+(2+2\lambda)=0\), i.e. \(8-\lambda=0\).
3. So \(\lambda=8\).

Teacher's Note:
1. Perpendicular vectors have zero dot product. Collect components carefully.

Answer: (b) perpendicular

Explanation:
1. The dot product is \(|\mathbf{a}|^2|\mathbf{b}|^2-|\mathbf{b}|^2|\mathbf{a}|^2=0\).
2. So the two vectors are perpendicular (they are the diagonals of a rhombus).

Teacher's Note:
1. Compute the dot product of the sum and difference; the cross terms cancel.

Answer: (a) Any vector

Explanation:
1. \(\mathbf{a}\cdot\mathbf{a}=|\mathbf{a}|^2=0\) gives \(\mathbf{a}=\mathbf{0}\).
2. Then \(\mathbf{a}\cdot\mathbf{b}=0\) holds for every \(\mathbf{b}\), so \(\mathbf{b}\) can be any vector.

Teacher's Note:
1. The zero vector is perpendicular to every vector.

Answer: (c) \(-\frac32\)

Explanation:
1. \(|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=0=3+2(\mathbf{a}\cdot\mathbf{b}+\mathbf{b}\cdot\mathbf{c}+\mathbf{c}\cdot\mathbf{a})\).
2. So the sum of the three dot products is \(-\frac32\).

Teacher's Note:
1. Square the given vector equation and use \(|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=1\).

Answer: (c) \(\cos^{-1}\frac{10}{\sqrt{102}}\)

Explanation:
1. \(\overrightarrow{BA}=(2,2,3)\) and \(\overrightarrow{BC}=(1,1,2)\).
2. \(\overrightarrow{BA}\cdot\overrightarrow{BC}=2+2+6=10\), \(|BA|=\sqrt{17}\), \(|BC|=\sqrt6\).
3. \(\cos\angle ABC=\frac{10}{\sqrt{102}}\).

Teacher's Note:
1. The angle at B uses the vectors that start at B.

Answer: (a) collinear

Explanation:
1. \(\overrightarrow{AB}=(1,4,-4)\) and \(\overrightarrow{BC}=(1,4,-4)\).
2. The two vectors are equal, so A, B, C lie on one line.

Teacher's Note:
1. If two position-difference vectors are parallel and share a point, the three points are collinear.

Answer: (b) \(|\mathbf{a}|=0\) or \(|\mathbf{b}|=0\)

Explanation:
1. \(\mathbf{a}\cdot\mathbf{b}=0\) means \(|\mathbf{a}||\mathbf{b}|\cos\theta=0\) and \(\mathbf{a}\times\mathbf{b}=\mathbf{0}\) means \(|\mathbf{a}||\mathbf{b}|\sin\theta=0\).
2. Both can hold only if \(|\mathbf{a}||\mathbf{b}|=0\), so at least one of them is a zero vector.

Teacher's Note:
1. \(\sin heta\) and \(\cos heta\) cannot both vanish.

Answer: (d) \(\frac{\sqrt3}{2}\hat i+\frac12\hat j\)

Explanation:
1. The unit vector at angle \(\theta\) is \(\cos\theta\,\hat i+\sin\theta\,\hat j\).
2. For \(\theta=30^\circ\) it is \(\frac{\sqrt3}{2}\hat i+\frac12\hat j\).

Teacher's Note:
1. Check that the magnitude of your answer is 1.

Answer: (c) \(\pm\left(\frac{3\sqrt{10}}{2}\hat i+\frac{\sqrt{10}}{2}\hat j\right)\)

Explanation:
1. The resultant is \(\mathbf{a}+\mathbf{b}=3\hat i+\hat j\), with magnitude \(\sqrt{10}\).
2. The required vector is \(\pm\frac{5}{\sqrt{10}}(3\hat i+\hat j)\).
3. \(\frac{5}{\sqrt{10}}=\frac{\sqrt{10}}{2}\), so it equals \(\pm\left(\frac{3\sqrt{10}}{2}\hat i+\frac{\sqrt{10}}{2}\hat j\right)\).

Teacher's Note:
1. Remember the \(\pm\): there are two vectors parallel to the resultant with the given magnitude.

Answer: (a) \(\frac37\hat i-\frac67\hat j+\frac27\hat k\) and \(11\sqrt5\) sq units

Explanation:
1. The diagonal is the sum: \(3\hat i-6\hat j+2\hat k\), of magnitude \(\sqrt{9+36+4}=7\). The unit vector is \(\frac37\hat i-\frac67\hat j+\frac27\hat k\).
2. The cross product of the sides is \(22\hat i+11\hat j+0\hat k\), of magnitude \(11\sqrt5\).
3. So the area is \(11\sqrt5\) sq units.

Teacher's Note:
1. Diagonal means sum of adjacent sides. Area means magnitude of their cross product.

Answer: (c) \(\frac53(32\hat i-\hat j-14\hat k)\)

Explanation:
1. \(\mathbf{a}\times\mathbf{b}=32\hat i-\hat j-14\hat k\), so \(\mathbf{d}=\lambda(32\hat i-\hat j-14\hat k)\).
2. \(\mathbf{c}\cdot\mathbf{d}=\lambda(64+1-56)=9\lambda=15\), so \(\lambda=\frac53\).
3. Hence \(\mathbf{d}=\frac53(32\hat i-\hat j-14\hat k)\).

Teacher's Note:
1. A vector perpendicular to two vectors is parallel to their cross product.

Answer: (a) 1

Explanation:
1. The sum is \((2+\lambda)\hat i+6\hat j-2\hat k\). Its dot product with \(\hat i+\hat j+\hat k\) is \(\lambda+6\).
2. Setting \(\frac{\lambda+6}{\sqrt{(2+\lambda)^2+36+4}}=1\): \((\lambda+6)^2=(\lambda+2)^2+40\).
3. \(8\lambda=8\), so \(\lambda=1\).

Teacher's Note:
1. Divide by the magnitude of the sum to make it a unit vector before taking the dot product.

Answer: (b) \(\Leftrightarrow\mathbf{a}\perp\mathbf{b}\)

Explanation:
1. \(|\mathbf{a}+\mathbf{b}|^2=|\mathbf{a}|^2+|\mathbf{b}|^2+2\mathbf{a}\cdot\mathbf{b}\).
2. Equality with \(|\mathbf{a}|^2+|\mathbf{b}|^2\) needs \(\mathbf{a}\cdot\mathbf{b}=0\), i.e. \(\mathbf{a}\perp\mathbf{b}\).

Teacher's Note:
1. This is the vector form of the Pythagoras theorem.

Answer: (b) \(0\le\theta\le\frac\pi2\)

Explanation:
1. \(\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta\) and \(\cos\theta\ge0\) for \(0\le\theta\le\frac\pi2\).
2. The end points are included because \(\cos0=1\) and \(\cos\frac\pi2=0\).

Teacher's Note:
1. Include the boundary cases where the vectors are parallel or perpendicular.

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

Explanation:
1. \(|\mathbf{a}+\mathbf{b}|^2=1+1+2\cos\theta=1\), so \(\cos\theta=-\frac12\).
2. Hence \(\theta=\frac{2\pi}{3}\).

Teacher's Note:
1. Set the squared magnitude equal to 1 and solve for \(\cos heta\).

Answer: (b) \(\frac\pi4\)

Explanation:
1. \(|\mathbf{a}||\mathbf{b}||\cos\theta|=|\mathbf{a}||\mathbf{b}||\sin\theta|\).
2. So \(|\tan\theta|=1\). Among the options, \(\theta=\frac\pi4\).

Teacher's Note:
1. Equal magnitudes of dot and cross products mean \(| an heta|=1\).

Answer: (d) 8

Explanation:
1. The direction ratios of the first and third points are \((4,12,4)\), i.e. \((1,3,1)\).
2. The first and second points give \((3,m+1,3)\), which must be \(3(1,3,1)\).
3. So \(m+1=9\) and \(m=8\).

Teacher's Note:
1. Collinear points give proportional direction ratios.