JEE Mathematics Vectors MCQs Set 32

Answer: (b) \(0\)

Explanation:
1. A non-zero vector perpendicular to all of \(\mathbf a,\mathbf b,\mathbf c\) exists only if the three vectors are coplanar.
2. So \([\mathbf a\ \mathbf b\ \mathbf c]=0\).
3. \([\mathbf a\times\mathbf b\ \ \mathbf b\times\mathbf c\ \ \mathbf c\times\mathbf a]=[\mathbf a\ \mathbf b\ \mathbf c]^2=0\).

Teacher's Note:
1. Use the identity \([\mathbf a\times\mathbf b\ \ \mathbf b\times\mathbf c\ \ \mathbf c\times\mathbf a]=[\mathbf a\ \mathbf b\ \mathbf c]^2\).
2. Key result: \([\mathbf a\times\mathbf b\ \ \mathbf b\times\mathbf c\ \ \mathbf c\times\mathbf a]=[\mathbf a\ \mathbf b\ \mathbf c]^2=0\).

Answer: (a) \(0\)

Explanation:
1. The triple product equals the determinant of the coefficients times \([\mathbf a\ \mathbf b\ \mathbf c]\).
2. Since \(\mathbf a,\mathbf b,\mathbf c\) are coplanar, \([\mathbf a\ \mathbf b\ \mathbf c]=0\), so the value is 0.

Teacher's Note:
1. Coplanar vectors have a zero scalar triple product, whatever the coefficients.
2. Key result: Since \(\mathbf a,\mathbf b,\mathbf c\) are coplanar, \([\mathbf a\ \mathbf b\ \mathbf c]=0\), so the value is 0.

Answer: (d) \(3\)

Explanation:
1. For this reciprocal system, \(\mathbf a\cdot\mathbf p=\mathbf b\cdot\mathbf q=\mathbf c\cdot\mathbf r=1\), and the other dot products such as \(\mathbf b\cdot\mathbf p\) are 0.
2. So the sum is \(1+1+1=3\).

Teacher's Note:
1. In a reciprocal system \(\mathbf a\cdot\mathbf p=1\) and \(\mathbf a\cdot\mathbf q=\mathbf a\cdot\mathbf r=0\), since a repeated vector in a triple product gives zero.
2. Key result: So the sum is \(1+1+1=3\).

Answer: (c) \(2\sqrt{26}\)

Explanation:
1. \(\mathbf a\cdot\mathbf b=2\sqrt2\cdot3\cos45^\circ=6\).
2. The diagonals are the sum \(3\mathbf a-2\mathbf b\) and the difference \(\mathbf a-4\mathbf b\). \(|3\mathbf a-2\mathbf b|^2=72+36-72=36\); \(|\mathbf a-4\mathbf b|^2=8+144-48=104\).
3. The longer diagonal is \(\sqrt{104}=2\sqrt{26}\).

Teacher's Note:
1. Find both diagonals as the sum and the difference of the sides, then compare lengths.
2. Key result: The longer diagonal is \(\sqrt{104}=2\sqrt{26}\).

Answer: (a) \(inclined at an obtuse angle\)

Explanation:
1. Squaring, \(|\mathbf a|^2+|\mathbf b|^2+2\mathbf a\cdot\mathbf b<|\mathbf a|^2+|\mathbf b|^2-2\mathbf a\cdot\mathbf b\), so \(\mathbf a\cdot\mathbf b<0\).
2. A negative dot product means the angle between them is obtuse.

Teacher's Note:
1. Compare the squares of the magnitudes; the dot product decides the sign.
2. Key result: A negative dot product means the angle between them is obtuse.

Answer: (a) \((\mathbf a\times\mathbf c)\times\mathbf b=\mathbf 0\)

Explanation:
1. \((\mathbf a\times\mathbf b)\times\mathbf c=\mathbf b(\mathbf a\cdot\mathbf c)-\mathbf a(\mathbf b\cdot\mathbf c)\) and \(\mathbf a\times(\mathbf b\times\mathbf c)=\mathbf b(\mathbf a\cdot\mathbf c)-\mathbf c(\mathbf a\cdot\mathbf b)\).
2. They are equal iff \(\mathbf a(\mathbf b\cdot\mathbf c)=\mathbf c(\mathbf a\cdot\mathbf b)\).
3. Now \((\mathbf a\times\mathbf c)\times\mathbf b=\mathbf c(\mathbf a\cdot\mathbf b)-\mathbf a(\mathbf b\cdot\mathbf c)\), which is zero in exactly that case.

Teacher's Note:
1. Expand both sides with the vector triple product formula and compare.
2. Key result: Now \((\mathbf a\times\mathbf c)\times\mathbf b=\mathbf c(\mathbf a\cdot\mathbf b)-\mathbf a(\mathbf b\cdot\mathbf c)\), which is zero in exactly that case.

Answer: (d) \((i) and (iv) only\)

Explanation:
1. \(|\overrightarrow{OA}|=5\) and \(|\overrightarrow{OB}|=15\). The bisector direction is \(\frac{\overrightarrow{OA}}{5}+\frac{\overrightarrow{OB}}{15}=\frac{(-12+14,\ 2,\ 9-5)}{15}=\frac{(2,2,4)}{15}\).
2. So the bisector ray is along \((1,1,2)\) with positive multiples.
3. \((2,2,4)\) and \((1,1,2)\) lie on it; \((-3,-3,-6)\) is on the opposite ray and \((2,11,5)\) is not on the line.

Teacher's Note:
1. Add the unit vectors along the two arms to get the internal bisector direction.
2. Key result: \((2,2,4)\) and \((1,1,2)\) lie on it; \((-3,-3,-6)\) is on the opposite ray and \((2,11,5)\) is not on the line.

Answer: (b) \(\frac\pi3\)

Explanation:
1. \(\mathbf c\cdot\mathbf d=5|\mathbf a|^2+6\,\mathbf a\cdot\mathbf b-8|\mathbf b|^2=5+6\,\mathbf a\cdot\mathbf b-8=0\).
2. So \(\mathbf a\cdot\mathbf b=\frac12\) and the angle is \(\frac\pi3\).

Teacher's Note:
1. Expand the dot product using \(|\mathbf a|=|\mathbf b|=1\).
2. Key result: So \(\mathbf a\cdot\mathbf b=\frac12\) and the angle is \(\frac\pi3\).

Answer: (a) \(-5\)

Explanation:
1. Both are unit vectors and \(\mathbf a\cdot\mathbf b=\frac{6-6}{7\sqrt{10}}=0\).
2. \((\mathbf a\times\mathbf b)\times(\mathbf a+2\mathbf b)=\mathbf b-2\mathbf a\) when \(\mathbf a\cdot\mathbf b=0\) and \(|\mathbf a|=|\mathbf b|=1\).
3. \((2\mathbf a-\mathbf b)\cdot(\mathbf b-2\mathbf a)=-|2\mathbf a-\mathbf b|^2=-(4+1)=-5\).

Teacher's Note:
1. Use the vector triple product formula and the fact that the two vectors are orthogonal unit vectors.
2. Key result: \((2\mathbf a-\mathbf b)\cdot(\mathbf b-2\mathbf a)=-|2\mathbf a-\mathbf b|^2=-(4+1)=-5\).

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

Explanation:
1. The Gram determinant must be non-negative: \(\begin{vmatrix}1&c&c\\c&1&c\\c&c&1\end{vmatrix}=(1-c)^2(1+2c)\ge0\) where \(c=\cos\theta\).
2. So \(\cos\theta\ge-\frac12\), hence \(\theta\le\frac{2\pi}{3}\).

Teacher's Note:
1. The square of a scalar triple product is the Gram determinant, so it cannot be negative.
2. Key result: So \(\cos\theta\ge-\frac12\), hence \(\theta\le\frac{2\pi}{3}\).

Answer: (a) \(-30\)

Explanation:
1. \([2\hat i\ \ 3\hat j\ \ -5\hat k]=2\cdot3\cdot(-5)[\hat i\ \hat j\ \hat k]=-30\).

Teacher's Note:
1. Scalar multiples come out of a triple product.
2. Key result: \([2\hat i\ \ 3\hat j\ \ -5\hat k]=2\cdot3\cdot(-5)[\hat i\ \hat j\ \hat k]=-30\).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A non-zero vector perpendicular to \(\mathbf a,\mathbf b,\mathbf c\) exists only if they are coplanar, so R is true.
2. The vectors \(\mathbf a-\mathbf b\), \(\mathbf b-\mathbf c\), \(\mathbf c-\mathbf a\) add to zero, so they are always coplanar and A is true for any \(\mathbf a,\mathbf b,\mathbf c\). R is therefore not the reason for A.

Teacher's Note:
1. The three difference vectors always sum to zero, whatever \(\mathbf a,\mathbf b,\mathbf c\) are.
2. Key result: The vectors \(\mathbf a-\mathbf b\), \(\mathbf b-\mathbf c\), \(\mathbf c-\mathbf a\) add to zero, so they are always coplanar and A is true for any \(\mathbf a,\mathbf b,\mathbf c\). R is therefore not the reason for A.

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

Explanation:
1. \(\mathbf b\times(\mathbf c-\mathbf d)=\mathbf 0\), so \(\mathbf c-\mathbf d=\lambda\mathbf b\), i.e. \(\mathbf d=\mathbf c-\lambda\mathbf b\).
2. \(\mathbf a\cdot\mathbf d=0\) gives \(\mathbf a\cdot\mathbf c=\lambda\,\mathbf a\cdot\mathbf b\), so \(\lambda=\frac{\mathbf a\cdot\mathbf c}{\mathbf a\cdot\mathbf b}\).
3. Hence \(\mathbf d=\mathbf c-\frac{\mathbf a\cdot\mathbf c}{\mathbf a\cdot\mathbf b}\mathbf b\).

Teacher's Note:
1. Parallel vectors have a zero cross product, so the difference is a multiple of the common vector.
2. Key result: Hence \(\mathbf d=\mathbf c-\frac{\mathbf a\cdot\mathbf c}{\mathbf a\cdot\mathbf b}\mathbf b\).

Answer: (a) \(\frac\pi3\)

Explanation:
1. \(\mathbf c=-(\mathbf a+\mathbf b)\), so \(49=9+25+2\mathbf a\cdot\mathbf b\), giving \(\mathbf a\cdot\mathbf b=\frac{15}{2}\).
2. \(\cos\theta=\frac{15/2}{15}=\frac12\), so \(\theta=\frac\pi3\).

Teacher's Note:
1. Square the relation \(\mathbf c=-(\mathbf a+\mathbf b)\) to get the dot product.
2. Key result: \(\cos\theta=\frac{15/2}{15}=\frac12\), so \(\theta=\frac\pi3\).

Answer: (d) \(\sqrt3\)

Explanation:
1. \(|\mathbf a+\mathbf b|^2=2+2\mathbf a\cdot\mathbf b=1\), so \(\mathbf a\cdot\mathbf b=-\frac12\).
2. \(|\mathbf a-\mathbf b|^2=2-2\mathbf a\cdot\mathbf b=3\), so \(|\mathbf a-\mathbf b|=\sqrt3\).

Teacher's Note:
1. Use \(|\mathbf a\pm\mathbf b|^2=|\mathbf a|^2+|\mathbf b|^2\pm2\mathbf a\cdot\mathbf b\).
2. Key result: \(|\mathbf a-\mathbf b|^2=2-2\mathbf a\cdot\mathbf b=3\), so \(|\mathbf a-\mathbf b|=\sqrt3\).

Answer: (d) \(\frac{5\pi}{6}\)

Explanation:
1. \(\mathbf b\cdot\mathbf c=-3|\mathbf b|^2=-48\), since \(\mathbf a\times\mathbf b\perp\mathbf b\).
2. \(|\mathbf a\times\mathbf b|^2=16-4=12\), so \(|\mathbf c|^2=4\cdot12+9\cdot16=192\) and \(|\mathbf c|=8\sqrt3\).
3. \(\cos\theta=\frac{-48}{4\cdot8\sqrt3}=-\frac{\sqrt3}{2}\), so \(\theta=\frac{5\pi}{6}\).

Teacher's Note:
1. The cross product term is perpendicular to \(\mathbf b\), so it adds nothing to the dot product.
2. Key result: \(\cos\theta=\frac{-48}{4\cdot8\sqrt3}=-\frac{\sqrt3}{2}\), so \(\theta=\frac{5\pi}{6}\).

Answer: (d) \(2\sqrt{14}\)

Explanation:
1. \(\hat i\times(\mathbf a\times\hat i)=\mathbf a(\hat i\cdot\hat i)-\hat i(\hat i\cdot\mathbf a)=\mathbf a-a_1\hat i\).
2. Adding the three terms, \(\mathbf b=3\mathbf a-\mathbf a=2\mathbf a\).
3. \(|\mathbf b|=2\sqrt{1+4+9}=2\sqrt{14}\).

Teacher's Note:
1. Apply \(\mathbf x\times(\mathbf y\times\mathbf z)=\mathbf y(\mathbf x\cdot\mathbf z)-\mathbf z(\mathbf x\cdot\mathbf y)\) to each term.
2. Key result: \(|\mathbf b|=2\sqrt{1+4+9}=2\sqrt{14}\).

Answer: (a) \(-1\)

Explanation:
1. \([\hat i\ \hat k\ \hat j]=-1\) (one swap of \([\hat i\ \hat j\ \hat k]=1\)), \([\hat k\ \hat j\ \hat i]=-1\), and \([\hat j\ \hat k\ \hat i]=+1\) (cyclic).
2. The sum is \(-1-1+1=-1\).

Teacher's Note:
1. A cyclic permutation keeps the sign of a triple product and a single swap reverses it.
2. Key result: The sum is \(-1-1+1=-1\).

Answer: (d) \(\frac{\hat j-\hat k}{\sqrt2}\)

Explanation:
1. A vector coplanar with the two given vectors is \(p(1,1,2)+q(1,2,1)\).
2. Perpendicular to \((1,1,1)\): \(4p+4q=0\), so \(q=-p\) and the vector is \(p(0,-1,1)\).
3. The unit vector is \(\pm\frac{\hat j-\hat k}{\sqrt2}\); the option listed is \(\frac{\hat j-\hat k}{\sqrt2}\).

Teacher's Note:
1. Write the coplanar vector as a combination of the two given vectors and impose the perpendicularity.
2. Key result: The unit vector is \(\pm\frac{\hat j-\hat k}{\sqrt2}\); the option listed is \(\frac{\hat j-\hat k}{\sqrt2}\).

Answer: (b) \(0\)

Explanation:
1. The coefficient determinant is \(\begin{vmatrix}1&-1&0\\0&1&1\\1&0&1\end{vmatrix}=1(1-0)+1(0-1)=0\).
2. Hence the triple product is \(0\cdot[\mathbf a\ \mathbf b\ \mathbf c]=0\).

Teacher's Note:
1. Convert each factor to coefficients of \(\mathbf a,\mathbf b,\mathbf c\) and expand a \(3\times3\) determinant.
2. Key result: Hence the triple product is \(0\cdot[\mathbf a\ \mathbf b\ \mathbf c]=0\).

Answer: (c) \(\frac34\)

Explanation:
1. \((\mathbf a+\lambda\mathbf b)\cdot(\mathbf a-\lambda\mathbf b)=|\mathbf a|^2-\lambda^2|\mathbf b|^2=0\).
2. So \(\lambda^2=\frac{9}{16}\) and \(\lambda=\pm\frac34\).

Teacher's Note:
1. Two vectors are perpendicular when their dot product is zero. The cross terms cancel.
2. Key result: So \(\lambda^2=\frac{9}{16}\) and \(\lambda=\pm\frac34\).

Answer: (c) \(\frac17(3\hat i+6\hat j-2\hat k)\)

Explanation:
1. The diagonal is \(\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=3\hat i+6\hat j-2\hat k\).
2. \(|\overrightarrow{AC}|=\sqrt{9+36+4}=7\).
3. The unit vector is \(\frac17(3\hat i+6\hat j-2\hat k)\).

Teacher's Note:
1. The diagonal of a parallelogram from a vertex is the sum of the two sides from that vertex.
2. Key result: The unit vector is \(\frac17(3\hat i+6\hat j-2\hat k)\).

Answer: (c) \(l^2+m^2+n^2\)

Explanation:
1. Expanding, the terms are \(l^2\,\mathbf a\cdot\mathbf p+lm\,\mathbf a\cdot\mathbf q+\dots\). Reciprocal vectors satisfy \(\mathbf a\cdot\mathbf p=\mathbf b\cdot\mathbf q=\mathbf c\cdot\mathbf r=1\) and all other products are 0.
2. The result is \(l^2+m^2+n^2\).

Teacher's Note:
1. The only surviving terms are those pairing a vector with its own reciprocal.
2. Key result: The result is \(l^2+m^2+n^2\).

Answer: (a) \(20\)

Explanation:
1. \(|\mathbf a\times\mathbf b|^2+(\mathbf a\cdot\mathbf b)^2=|\mathbf a|^2|\mathbf b|^2\).
2. \(16+4=20\).

Teacher's Note:
1. Combine \(|\mathbf a||\mathbf b|\sin\theta\) and \(|\mathbf a||\mathbf b|\cos\theta\) by squaring and adding.
2. Key result: \(16+4=20\).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \(|\mathbf a|=3\) and \(|\mathbf b|=9\), so \(\frac{\mathbf a}{3}+\frac{\mathbf b}{9}=\frac{3\mathbf a+\mathbf b}{9}=\frac{(-5,7,2)}{9}\), which is along \(\mathbf c\). Statement A is true.
2. Also \(\frac{\mathbf c\cdot\mathbf a}{|\mathbf a|}=\frac{13}{3}\) and \(\frac{\mathbf c\cdot\mathbf b}{|\mathbf b|}=\frac{39}{9}=\frac{13}{3}\), so the inclinations are equal. R is true and it explains A, since the bisector is exactly the direction equally inclined to both.

Teacher's Note:
1. The internal bisector of two vectors is the direction of the sum of the unit vectors.
2. Key result: Also \(\frac{\mathbf c\cdot\mathbf a}{|\mathbf a|}=\frac{13}{3}\) and \(\frac{\mathbf c\cdot\mathbf b}{|\mathbf b|}=\frac{39}{9}=\frac{13}{3}\), so the inclinations are equal. R is true and it explains A, since the bisector is exactly the direction equally inclined to both.

Answer: (d) \(4\)

Explanation:
1. The projection of \(\mathbf a\) on \(\mathbf b\) is \(|\mathbf a|\cos\frac{5\pi}{6}=-\frac{\sqrt3}{2}|\mathbf a|\).
2. Setting \(-\frac{\sqrt3}{2}|\mathbf a|=-\frac{6}{\sqrt3}=-2\sqrt3\) gives \(|\mathbf a|=4\).

Teacher's Note:
1. Projection of a on b equals \(|\mathbf a|\cos\theta\), which is negative for an obtuse angle.
2. Key result: Setting \(-\frac{\sqrt3}{2}|\mathbf a|=-\frac{6}{\sqrt3}=-2\sqrt3\) gives \(|\mathbf a|=4\).

Answer: (d) \(none of these\)

Explanation:
1. Let the unit vector be \((\cos\theta,\sin\theta)\). Making \(45^\circ\) with \(\hat i+\hat j\) (direction \(45^\circ\)) requires \(\theta=0\) or \(90^\circ\).
2. For \(\theta=0\): \(\cos\) of the angle with \(3\hat i-4\hat j\) is \(\frac35\ne\frac12\). For \(\theta=90^\circ\): it is \(-\frac45\ne\frac12\).
3. No such vector exists among those candidates, so none of the options is correct.

Teacher's Note:
1. Reduce the first condition to two candidates and test them against the second condition.
2. Key result: No such vector exists among those candidates, so none of the options is correct.

Answer: (c) \(\hat i\)

Explanation:
1. \(\mathbf a\times(\mathbf a\times\mathbf b)=\mathbf a(\mathbf a\cdot\mathbf b)-\mathbf b|\mathbf a|^2\). The left side is \((1,1,1)\times(0,1,-1)=(-2,1,1)\).
2. So \((1,1,1)-3\mathbf b=(-2,1,1)\), giving \(3\mathbf b=(3,0,0)\).
3. Hence \(\mathbf b=\hat i\). Check: \(\mathbf a\cdot\mathbf b=1\) and \(\mathbf a\times\hat i=\hat j-\hat k\).

Teacher's Note:
1. Cross the given relation with \(\mathbf a\) and expand the vector triple product.
2. Key result: Hence \(\mathbf b=\hat i\). Check: \(\mathbf a\cdot\mathbf b=1\) and \(\mathbf a\times\hat i=\hat j-\hat k\).

Answer: (a) \(\cos^{-1}\left(\frac{4}{\sqrt{21}}\right)\)

Explanation:
1. \(|\mathbf x|^2-1=8\), so \(|\mathbf x|^2=9\).
2. \(|\mathbf x-\hat{\mathbf a}|^2=9+1-4=6\) and \(|\mathbf x+\hat{\mathbf a}|^2=9+1+4=14\).
3. \(\cos\theta=\frac{8}{\sqrt6\sqrt{14}}=\frac{8}{2\sqrt{21}}=\frac{4}{\sqrt{21}}\).

Teacher's Note:
1. Find \(|\mathbf x|\) first, then both lengths.
2. Key result: \(\cos\theta=\frac{8}{\sqrt6\sqrt{14}}=\frac{8}{2\sqrt{21}}=\frac{4}{\sqrt{21}}\).

Answer: (d) \(3\)

Explanation:
1. Vector from the point on the line to the given point: \((3,0,5)-(8,3,1)=(-5,-3,4)\). Direction \(\mathbf d=(6,1,-2)\).
2. \((-5,-3,4)\times(6,1,-2)=(2,14,13)\), of magnitude \(\sqrt{4+196+169}=\sqrt{369}\). Also \(|\mathbf d|=\sqrt{41}\).
3. Distance \(=\sqrt{\frac{369}{41}}=3\).

Teacher's Note:
1. Distance from a point to a line is \(\frac{|\overrightarrow{PA}\times\mathbf d|}{|\mathbf d|}\).
2. Key result: Distance \(=\sqrt{\frac{369}{41}}=3\).