JEE Mathematics Vectors MCQs Set 33

Answer: (b) \(3\overrightarrow{AC}\)

Explanation:
1. \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\).
2. \(\overrightarrow{AE}+\overrightarrow{ED}+\overrightarrow{DC}=\overrightarrow{AC}\).
3. Adding the extra \(\overrightarrow{AC}\), the total is \(3\overrightarrow{AC}\).

Teacher's Note:
1. Group the vectors into chains that start at A and end at C.
2. Key result: Adding the extra \(\overrightarrow{AC}\), the total is \(3\overrightarrow{AC}\).

Answer: (d) \(a=\frac{1}{|\lambda|}\)

Explanation:
1. \(|\lambda\mathbf a|=|\lambda|\,a=1\), so \(a=\frac{1}{|\lambda|}\).

Teacher's Note:
1. The magnitude of a scalar multiple is \(|\lambda|\) times the magnitude.
2. Key result: \(|\lambda\mathbf a|=|\lambda|\,a=1\), so \(a=\frac{1}{|\lambda|}\).

Answer: (c) \(-7\)

Explanation:
1. \(|\mathbf a+\mathbf b+\mathbf c|^2=0\) gives \(1+4+9+2S=0\), so \(S=-7\).

Teacher's Note:
1. The same squaring trick gives \(S=-\frac{|\mathbf a|^2+|\mathbf b|^2+|\mathbf c|^2}{2}\).
2. Key result: \(|\mathbf a+\mathbf b+\mathbf c|^2=0\) gives \(1+4+9+2S=0\), so \(S=-7\).

Answer: (d) \(90^\circ\)

Explanation:
1. \(\mathbf a+\mathbf b=(2,0,8)\) and \(\mathbf a-\mathbf b=(0,2,0)\).
2. Their dot product is \(0\), so the angle is \(90^\circ\).

Teacher's Note:
1. Compute both vectors and test the dot product.
2. Key result: Their dot product is \(0\), so the angle is \(90^\circ\).

Answer: (c) \(\alpha=\pm1,\ \beta=1\)

Explanation:
1. Linear dependence means \(\begin{vmatrix}1&1&1\\4&3&4\\1&\alpha&\beta\end{vmatrix}=1-\beta=0\), so \(\beta=1\).
2. \(|\mathbf c|^2=1+\alpha^2+1=3\), so \(\alpha=\pm1\).

Teacher's Note:
1. Set the determinant of the three vectors equal to zero.
2. Key result: \(|\mathbf c|^2=1+\alpha^2+1=3\), so \(\alpha=\pm1\).

Answer: (a) \((2,-1,-1)\)

Explanation:
1. Comparing components: \(\alpha+2\beta+3\gamma=-3\), \(2\alpha+3\beta+\gamma=0\), \(3\alpha+\beta+2\gamma=3\).
2. Solving, \(\alpha=2\), \(\beta=-1\), \(\gamma=-1\). Check: \(2-2-3=-3\), \(4-3-1=0\), \(6-1-2=3\).

Teacher's Note:
1. Write the vector equation as three linear equations and solve.
2. Key result: Solving, \(\alpha=2\), \(\beta=-1\), \(\gamma=-1\). Check: \(2-2-3=-3\), \(4-3-1=0\), \(6-1-2=3\).

Answer: (c) \(5\sqrt2\)

Explanation:
1. \(\mathbf a\cdot(\mathbf b+\mathbf c)=0\), \(\mathbf b\cdot(\mathbf a+\mathbf c)=0\), \(\mathbf c\cdot(\mathbf a+\mathbf b)=0\). Adding, \(2(\mathbf a\cdot\mathbf b+\mathbf b\cdot\mathbf c+\mathbf c\cdot\mathbf a)=0\).
2. \(|\mathbf a+\mathbf b+\mathbf c|^2=9+16+25+0=50\), so the magnitude is \(5\sqrt2\).

Teacher's Note:
1. Add the three given conditions; the sum of the pairwise dot products is zero.
2. Key result: \(|\mathbf a+\mathbf b+\mathbf c|^2=9+16+25+0=50\), so the magnitude is \(5\sqrt2\).

Answer: (c) \(all of the above\)

Explanation:
1. The third vertex only needs to be off the line through the two given points. That line has direction \((-1,2,1)\).
2. Each of \(\hat i+\hat k\), \(\hat i-2\hat j-\hat k\) and \(-2\hat i-\hat j\) differs from \(\hat i-\hat j\) by a vector that is not parallel to \((-1,2,1)\), so all three give a triangle.

Teacher's Note:
1. Three points form a triangle exactly when they are not collinear.
2. Key result: Each of \(\hat i+\hat k\), \(\hat i-2\hat j-\hat k\) and \(-2\hat i-\hat j\) differs from \(\hat i-\hat j\) by a vector that is not parallel to \((-1,2,1)\), so all three give a triangle.

Answer: (d) \(\frac{1}{5\sqrt3}(5\hat i-\hat j+7\hat k)\)

Explanation:
1. Two edge vectors are \((-1,2,1)\) and \((2,3,-1)\). Their cross product is \((-5,1,-7)\), of magnitude \(\sqrt{75}=5\sqrt3\).
2. A unit normal is \(\pm\frac{1}{5\sqrt3}(5\hat i-\hat j+7\hat k)\).

Teacher's Note:
1. The normal's direction is fixed up to sign by the cross product of two edges.
2. Key result: A unit normal is \(\pm\frac{1}{5\sqrt3}(5\hat i-\hat j+7\hat k)\).

Answer: (c) \(x=\frac12,\ y=4,\ z=-10\)

Explanation:
1. Collinear vectors have proportional components: \(\frac x1=\frac2y=\frac{5}{-z}\).
2. For \(x=\frac12\): \(y=4\) and \(-z=10\), so \(z=-10\). This satisfies all three ratios equal to \(\frac12\).
3. The other sets break the proportion (for example \(x=1\) needs \(y=2\)).

Teacher's Note:
1. Test each set in the ratio condition.
2. Key result: The other sets break the proportion (for example \(x=1\) needs \(y=2\)).

Answer: (c) \(12\sqrt3\)

Explanation:
1. \(\mathbf a\) is perpendicular to the plane of \(\mathbf b,\mathbf c\), so \([\mathbf a\ \mathbf b\ \mathbf c]=|\mathbf a||\mathbf b\times\mathbf c|\).
2. \(|\mathbf b\times\mathbf c|=3\cdot4\sin\frac{2\pi}{3}=6\sqrt3\), so the value is \(2\cdot6\sqrt3=12\sqrt3\).

Teacher's Note:
1. A vector perpendicular to the plane of two others is parallel to their cross product.
2. Key result: \(|\mathbf b\times\mathbf c|=3\cdot4\sin\frac{2\pi}{3}=6\sqrt3\), so the value is \(2\cdot6\sqrt3=12\sqrt3\).

Answer: (b) \(neither x nor y\)

Explanation:
1. \(\begin{vmatrix}1&0&-1\\x&1&1-x\\y&x&1+x-y\end{vmatrix}=(1+x-y-x+x^2)-(x^2-y)=1\).
2. The value is the constant 1, independent of x and y.

Teacher's Note:
1. Expand the determinant carefully and watch the terms cancel.
2. Key result: The value is the constant 1, independent of x and y.

Answer: (a) \(4\overrightarrow{OM}\)

Explanation:
1. M is the midpoint of both AC and BD, so \(\overrightarrow{OA}+\overrightarrow{OC}=2\overrightarrow{OM}\) and \(\overrightarrow{OB}+\overrightarrow{OD}=2\overrightarrow{OM}\).
2. The sum is \(4\overrightarrow{OM}\).

Teacher's Note:
1. Pair opposite vertices through the midpoint of the diagonals.
2. Key result: The sum is \(4\overrightarrow{OM}\).

Answer: (c) \(-10\)

Explanation:
1. \(\begin{vmatrix}2&1&3\\-1&2&1\\3&1&2\end{vmatrix}=2(4-1)-1(-2-3)+3(-1-6)=6+5-21=-10\).

Teacher's Note:
1. The scalar triple product is the determinant of the three component rows.
2. Key result: \(\begin{vmatrix}2&1&3\\-1&2&1\\3&1&2\end{vmatrix}=2(4-1)-1(-2-3)+3(-1-6)=6+5-21=-10\).

Answer: (a) \(\frac{3\sqrt{26}}{7}\)

Explanation:
1. \(\overrightarrow{BA}=(-1,3,0)\), \(\overrightarrow{BC}=(-2,-6,3)\), \(|\overrightarrow{BC}|=7\).
2. \(\overrightarrow{BA}\times\overrightarrow{BC}=(9,3,12)\), of magnitude \(\sqrt{234}=3\sqrt{26}\).
3. Distance \(=\frac{3\sqrt{26}}{7}\).

Teacher's Note:
1. Distance of a point from a line is \(\frac{|\overrightarrow{BA}\times\overrightarrow{BC}|}{|\overrightarrow{BC}|}\).
2. Key result: Distance \(=\frac{3\sqrt{26}}{7}\).

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

Explanation:
1. \(|\cos\theta|=|\sin\theta|\), so \(|\tan\theta|=1\).
2. Among the listed values, \(\theta=\frac\pi4\) satisfies it (\(0\), \(\frac\pi2\) and \(\pi\) do not).

Teacher's Note:
1. Divide the two sides to get a condition on \(\tan\theta\).
2. Key result: Among the listed values, \(\theta=\frac\pi4\) satisfies it (\(0\), \(\frac\pi2\) and \(\pi\) do not).

Answer: (c) \(\sqrt{34}:\sqrt{45}:39\)

Explanation:
1. \(|\mathbf a|=\sqrt{9+25}=\sqrt{34}\) and \(|\mathbf b|=\sqrt{36+9}=\sqrt{45}\).
2. \(\mathbf a\times\mathbf b=(9+30)\hat k=39\hat k\), so \(|\mathbf c|=39\).

Teacher's Note:
1. Both vectors lie in the xy-plane, so their cross product is along \(\hat k\).
2. Key result: \(\mathbf a\times\mathbf b=(9+30)\hat k=39\hat k\), so \(|\mathbf c|=39\).

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

Explanation:
1. Cross \(\mathbf a+\mathbf b+\mathbf c=\mathbf 0\) with \(\mathbf a\): \(\mathbf a\times\mathbf b+\mathbf a\times\mathbf c=\mathbf 0\), so \(\mathbf a\times\mathbf b=\mathbf c\times\mathbf a\). Similarly the others are equal.
2. They are non-zero because three unit vectors summing to zero cannot be collinear (they are at \(120^\circ\)).

Teacher's Note:
1. Cross the relation with each vector in turn.
2. Key result: They are non-zero because three unit vectors summing to zero cannot be collinear (they are at \(120^\circ\)).

Answer: (b) \(adjacent sides of a rectangle\)

Explanation:
1. Squaring, \(4\,\mathbf a\cdot\mathbf b=0\), so \(\mathbf a\perp\mathbf b\).
2. Equal diagonals in a parallelogram means it is a rectangle; nothing says \(|\mathbf a|=|\mathbf b|\), so it need not be a square.

Teacher's Note:
1. The diagonals of a parallelogram are \(\mathbf a\pm\mathbf b\); they are equal exactly for a rectangle.
2. Key result: Equal diagonals in a parallelogram means it is a rectangle; nothing says \(|\mathbf a|=|\mathbf b|\), so it need not be a square.

Answer: (b) \(-24\hat i+32\hat j+30\hat k\)

Explanation:
1. \(|\mathbf b|=\sqrt{36+64+\frac{225}{4}}=\frac{25}{2}\), so \(\mathbf a=\pm\frac{50}{25/2}\mathbf b=\pm4\mathbf b\).
2. An acute angle with the positive z-axis needs a positive z-component: \(\mathbf a=-4\mathbf b=-24\hat i+32\hat j+30\hat k\).

Teacher's Note:
1. Scale \(\mathbf b\) to the required magnitude and choose the sign from the z-component.
2. Key result: An acute angle with the positive z-axis needs a positive z-component: \(\mathbf a=-4\mathbf b=-24\hat i+32\hat j+30\hat k\).

Answer: (c) \(-\frac{1}{\sqrt{165}}(10\hat i+7\hat j-4\hat k)\)

Explanation:
1. \(\overrightarrow{AB}=(-2,0,-5)\) and \(\overrightarrow{AC}=(1,-2,-1)\).
2. \(\overrightarrow{AB}\times\overrightarrow{AC}=(-10,-7,4)\), with magnitude \(\sqrt{165}\).
3. A unit normal is \(\pm\frac{(-10,-7,4)}{\sqrt{165}}=-\frac{1}{\sqrt{165}}(10\hat i+7\hat j-4\hat k)\).

Teacher's Note:
1. A normal to the plane is the cross product of two edges.
2. Key result: A unit normal is \(\pm\frac{(-10,-7,4)}{\sqrt{165}}=-\frac{1}{\sqrt{165}}(10\hat i+7\hat j-4\hat k)\).

Answer: (b) \(\pi\)

Explanation:
1. \(\mathbf a\) is a positive multiple of \(\mathbf b\) and \(\mathbf c\) a negative multiple, so they point in opposite directions.
2. The angle between them is \(\pi\).

Teacher's Note:
1. Parallel vectors with opposite signs of the scalar multiples are antiparallel.
2. Key result: The angle between them is \(\pi\).

Answer: (d) \(\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{CA}=\mathbf 0\)

Explanation:
1. By the triangle law \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\), so options 1, 2 and 4 are true (since \(-\overrightarrow{CB}=\overrightarrow{BC}\)).
2. Option 3 gives \(\overrightarrow{AC}-\overrightarrow{CA}=2\overrightarrow{AC}\ne\mathbf 0\), so it is not true.

Teacher's Note:
1. Apply the triangle law to each option; reversing a vector changes its sign.
2. Key result: Option 3 gives \(\overrightarrow{AC}-\overrightarrow{CA}=2\overrightarrow{AC}\ne\mathbf 0\), so it is not true.

Answer: (c) \(13\)

Explanation:
1. \((\mathbf a\times\mathbf b)^2+(\mathbf a\cdot\mathbf b)^2=|\mathbf a|^2|\mathbf b|^2=676\).
2. \(|\mathbf a|^2=\frac{676}{4}=169\), so \(|\mathbf a|=13\).

Teacher's Note:
1. Lagrange's identity: \(|\mathbf a\times\mathbf b|^2+(\mathbf a\cdot\mathbf b)^2=|\mathbf a|^2|\mathbf b|^2\).
2. Key result: \(|\mathbf a|^2=\frac{676}{4}=169\), so \(|\mathbf a|=13\).

Answer: (b) \(16\)

Explanation:
1. \(|\mathbf a\times\mathbf b|^2=|\mathbf a|^2|\mathbf b|^2-(\mathbf a\cdot\mathbf b)^2=400-144=256\).
2. So \(|\mathbf a\times\mathbf b|=16\).

Teacher's Note:
1. Lagrange's identity gives the magnitude directly.
2. Key result: So \(|\mathbf a\times\mathbf b|=16\).

Answer: (b) \(collinear\)

Explanation:
1. \(\overrightarrow{AB}=(1,-4,-2)\) and \(\overrightarrow{AC}=(-3,12,6)=-3\,\overrightarrow{AB}\).
2. The two vectors are parallel, so the points are collinear.

Teacher's Note:
1. Check whether one displacement is a scalar multiple of the other.
2. Key result: The two vectors are parallel, so the points are collinear.

Answer: (c) \(Q is a fixed point\)

Explanation:
1. Let M be the midpoint of BC. Then \(\overrightarrow{PB}+\overrightarrow{PC}=2\overrightarrow{PM}\).
2. \(\overrightarrow{AQ}=\overrightarrow{AP}+\overrightarrow{PQ}=2\overrightarrow{AP}+2\overrightarrow{PM}=2\overrightarrow{AM}\), which does not depend on P.
3. So Q is a fixed point.

Teacher's Note:
1. Use the midpoint of BC to combine \(\overrightarrow{PB}+\overrightarrow{PC}\).
2. Key result: So Q is a fixed point.

Answer: (a) \(3 sq. units\)

Explanation:
1. \(|\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{4+16+16}=6\).
2. Area of the triangle is half of this: \(3\).

Teacher's Note:
1. Triangle area is half the magnitude of the cross product of two sides.
2. Key result: Area of the triangle is half of this: \(3\).

Answer: (c) \(2\mathbf a^2\)

Explanation:
1. \((\mathbf a\times\hat i)^2=|\mathbf a|^2-a_1^2\), and similarly for the others.
2. The sum is \(3|\mathbf a|^2-|\mathbf a|^2=2|\mathbf a|^2\).

Teacher's Note:
1. Use \(|\mathbf a\times\mathbf u|^2=|\mathbf a|^2-(\mathbf a\cdot\mathbf u)^2\) for a unit vector \(\mathbf u\).
2. Key result: The sum is \(3|\mathbf a|^2-|\mathbf a|^2=2|\mathbf a|^2\).

Answer: (c) \(25\)

Explanation:
1. Squaring \(\mathbf a+\mathbf b+\mathbf c=\mathbf 0\): \(25+16+9+2(\mathbf a\cdot\mathbf b+\mathbf b\cdot\mathbf c+\mathbf c\cdot\mathbf a)=0\).
2. The sum is \(-25\), so its absolute value is \(25\).

Teacher's Note:
1. Square the zero-sum relation to find the sum of the pairwise dot products.
2. Key result: The sum is \(-25\), so its absolute value is \(25\).