JEE Mathematics Vector Algebra MCQs Set 10

Answer: (a) \( \dfrac{\vec{a}+\vec{a}\times\vec{c}+2\,\vec{a}\times\vec{b}}{3} \)

Explanation:
1. Take \( \vec{a}\times \) of \( \vec{u}+\vec{v}+\vec{w}=\vec{a} \) and use \( \vec{a}\times\vec{u}=\vec{a}\times\vec{a}-\vec{a}\times\vec{v}-\vec{a}\times\vec{w} \).
2. Solving the three relations (checked on random vectors satisfying the given conditions) gives \( \vec{u}=\dfrac{\vec{a}+\vec{a}\times\vec{c}+2\,\vec{a}\times\vec{b}}{3} \), \( \vec{v}=\dfrac{\vec{a}+\vec{a}\times\vec{c}-\vec{a}\times\vec{b}}{3} \), \( \vec{w}=\dfrac{\vec{a}-2\,\vec{a}\times\vec{c}-\vec{a}\times\vec{b}}{3} \).
3. These three add up to \( \vec{a} \), as required. Answer: option (a).

Teacher's Note:
1. Check any candidate by adding the three vectors; the sum must be \( \vec{a} \).
2. The coefficients of the cross terms must also cancel in the sum.

Answer: (c) \( \dfrac{\vec{a}+\vec{a}\times\vec{c}-\vec{a}\times\vec{b}}{3} \)

Explanation:
1. Solving the relations (verified numerically with random \( \vec{u},\vec{v} \) satisfying \( \vec{a}\cdot\vec{u}=\vec{a}\cdot\vec{v}=1 \)) gives \( \vec{v}=\dfrac{\vec{a}+\vec{a}\times\vec{c}-\vec{a}\times\vec{b}}{3} \).
2. Check: \( \vec{a}\cdot\vec{v}=\dfrac{3+0-0}{3}=1 \), as given. Answer: option (c).

Teacher's Note:
1. The scalar condition \( \vec{a}\cdot\vec{v}=1 \) is a quick test: cross products with \( \vec{a} \) are perpendicular to \( \vec{a} \).
2. Only the \( \vec{a}/3 \) part contributes, since \( |\vec{a}|^2=3 \).

Answer: (b) \( \dfrac{6}{\sqrt{14}} \)

Explanation:
1. The line has direction \( \hat{i}+2\hat{j}+3\hat{k} \) (coefficients of \( t \)), of magnitude \( \sqrt{14} \).
2. Projection \( =\dfrac{(1)(1)+(1)(2)+(1)(3)}{\sqrt{14}}=\dfrac{6}{\sqrt{14}} \). Answer: option (b).

Teacher's Note:
1. Read the direction from the coefficients of the parameter.
2. Divide the dot product by the magnitude of the direction vector, not by its square.

Answer: (d) \( \sqrt{5} \)

Explanation:
1. \( |\overrightarrow{AB}|=|\overrightarrow{AC}|=\sqrt{11} \), so the triangle is isosceles and the point of \( BC \) equidistant from \( AB \) and \( AC \) lies on the bisector of angle \( A \), which is the midpoint of \( BC \).
2. \( \overrightarrow{AP}=\tfrac12(\overrightarrow{AB}+\overrightarrow{AC})=2\hat{i}+\hat{k} \).
3. \( |\overrightarrow{AP}|=\sqrt{4+0+1}=\sqrt{5} \). Answer: option (d).

Teacher's Note:
1. Equal distances from the two arms means P lies on the angle bisector.
2. In an isosceles triangle the bisector from the apex meets the base at its midpoint.

Answer: (c) \( 13 \)

Explanation:
1. A point of the line is \( (2+3t,\,-1+4t,\,2+12t) \). Put it in the plane: \( (2+3t)-(-1+4t)+(2+12t)=5 \), so \( 5+11t=5 \) and \( t=0 \).
2. The intersection point is \( (2,-1,2) \).
3. Its distance from \( (-1,-5,-10) \) is \( \sqrt{9+16+144}=13 \). Answer: option (c).

Teacher's Note:
1. Substitute the parametric point of the line into the plane equation to find the parameter.
2. Then use the distance formula between two points.

Answer: (d) \( 0 \)

Explanation:
1. The line passes through \( P_0(-1,3,-2) \) with direction \( \vec{d}=(-3,2,1) \), and the given point is \( Q(0,7,-7) \).
2. Normal \( \vec{n}=\vec{d}\times\overrightarrow{P_0Q}=(-3,2,1)\times(1,4,-5)=(-14,-14,-14) \), which is parallel to \( (1,1,1) \).
3. The plane is \( x+y+z=k \); \( Q \) gives \( 0+7-7=0 \), so \( k=0 \) and the plane passes through the origin.
4. Distance \( =0 \). Answer: option (d).

Teacher's Note:
1. Find the normal from the line direction and a vector joining two known points of the plane.
2. Substituting a known point into \( x+y+z=k \) gives the constant at once.

Answer: (b) \( \sqrt{\dfrac{17}{2}} \)

Explanation:
1. Let \( P=(1,0,1) \), the line pass through \( A=(1,2,4) \) and have direction \( \vec{u}=(1,0,-1) \). Then \( \overrightarrow{AP}=(0,-2,-3) \).
2. \( \overrightarrow{AP}\times\vec{u}=(2,-3,2) \), magnitude squared \( =17 \).
3. Distance \( =\dfrac{|\overrightarrow{AP}\times\vec{u}|}{|\vec{u}|}=\dfrac{\sqrt{17}}{\sqrt{2}}=\sqrt{\dfrac{17}{2}} \). Answer: option (b).

Teacher's Note:
1. Distance of a point from a line is the magnitude of the cross product divided by the length of the direction vector.
2. Do not forget to divide by \( |\vec{u}| \).

Answer: (a) \( \dfrac{1}{2}\sqrt{3\sqrt{3}-5} \)

Explanation:
1. The volume is the square root of the Gram determinant \( 1-3\cos^2 30^\circ+2\cos^3 30^\circ \).
2. With \( \cos30^\circ=\dfrac{\sqrt3}{2} \): \( 1-\dfrac94+\dfrac{3\sqrt3}{4}=\dfrac{3\sqrt3-5}{4} \).
3. Volume \( =\dfrac12\sqrt{3\sqrt3-5} \). Answer: option (a).

Teacher's Note:
1. For three unit vectors with equal mutual angle, use \( [\vec a\,\vec b\,\vec c]^2=1-3\cos^2\theta+2\cos^3\theta \).
2. This is the determinant of the matrix of dot products.

Answer: (d) \( 0 \)

Explanation:
1. Since \( \vec{a},\vec{b},\vec{a}\times\vec{b} \) are mutually perpendicular unit vectors, \( \vec{c}\cdot\vec{a}=\lambda \) and \( \vec{c}\cdot\vec{b}=\mu \).
2. \( (\vec{a}-\vec{c})\cdot(\vec{b}+\vec{c})=0+\lambda-\mu-1=0 \), so \( \lambda=\mu+1 \).
3. \( |\vec{c}|=1 \) gives \( \lambda^2+\mu^2+\omega^2=1 \), i.e. \( (\mu+1)^2+\mu^2+\omega^2=1 \), so \( 2\mu^2+2\mu+\omega^2=0 \). Answer: option (d).

Teacher's Note:
1. Expand the unknown vector in the orthonormal triad \( \vec a,\vec b,\vec a\times\vec b \).
2. Two scalar conditions are enough here, and the required expression appears directly.

Answer: (b) \( \{(\vec{a}\times\vec{b})\cdot(\vec{a}\times\vec{b})\}\vec{c} \)

Explanation:
1. Since \( \vec{c} \) lies in the plane of \( \vec{a} \) and \( \vec{b} \), write \( \vec{c}=x\vec{a}+y\vec{b} \). Then \( \vec{c}\times\vec{b}=x\,\vec{a}\times\vec{b} \) and \( \vec{a}\times\vec{c}=y\,\vec{a}\times\vec{b} \).
2. The expression becomes \( |\vec a\times\vec b|^2(x\vec{a}+y\vec{b})=|\vec a\times\vec b|^2\,\vec{c} \) (also verified numerically). Answer: option (b).

Teacher's Note:
1. Coplanarity lets you write one vector as a combination of the other two.
2. Every cross product then becomes a scalar multiple of \( \vec a\times\vec b \).

Answer: (c) \( -9\vec{a} \)

Explanation:
1. Let \( 3\vec{a}+7\vec{b}=\lambda\vec{c} \) and \( 3\vec{b}+2\vec{c}=\mu\vec{a} \). Then \( \vec{c}=\dfrac{\mu\vec{a}-3\vec{b}}{2} \).
2. Substituting: \( 3\vec{a}+7\vec{b}=\dfrac{\lambda\mu}{2}\vec{a}-\dfrac{3\lambda}{2}\vec{b} \). As \( \vec a,\vec b \) are independent, \( \dfrac{\lambda\mu}{2}=3 \) and \( -\dfrac{3\lambda}{2}=7 \), so \( \lambda=-\dfrac{14}{3} \) and \( \mu=-\dfrac{9}{7} \).
3. \( 21\vec{b}+14\vec{c}=7(3\vec{b}+2\vec{c})=7\mu\vec{a}=-9\vec{a} \). Answer: option (c).

Teacher's Note:
1. Turn each collinearity statement into an equation with a scalar multiplier.
2. Compare coefficients of two non-collinear vectors on both sides.

Answer: (b) \( 3 \)

Explanation:
1. Take \( A \) as origin, \( \overrightarrow{AB}=\vec{p} \), \( \overrightarrow{AD}=\vec{q} \). Then \( C=\vec p+\vec q \), \( E=\vec p+\tfrac45\vec q \), \( F=\tfrac15\vec p+\vec q \).
2. A point of \( EF \): \( E+k(F-E)=(1-\tfrac45k)\vec p+(\tfrac45+\tfrac15k)\vec q \).
3. For it to lie on \( AC \) the two coefficients must be equal: \( 1-\tfrac45k=\tfrac45+\tfrac k5 \), giving \( k=\tfrac15 \).
4. So \( \lambda=1-\tfrac{4}{25}=\tfrac{21}{25} \) and \( \dfrac{25\lambda}{7}=3 \). Answer: option (b).

Teacher's Note:
1. Choose two adjacent sides as base vectors and write every point in terms of them.
2. A point lies on the diagonal when its two coefficients are equal.

Answer: (a) \( 5 \)

Explanation:
1. \( \overrightarrow{BA}=(3,0,0) \) and \( \vec{u}=(2,3,6) \), \( |\vec{u}|=7 \).
2. \( \overrightarrow{BA}\times\vec{u}=(0,-18,9) \), so \( |\overrightarrow{BA}\times\vec{u}|^2=405 \).
3. \( \lambda^2=\dfrac{405}{49} \), hence \( \dfrac{49\lambda^2}{81}=\dfrac{405}{81}=5 \). Answer: option (a).

Teacher's Note:
1. Distance from a point to a line equals \( |\overrightarrow{BA}\times\vec{u}|/|\vec{u}| \).
2. Square it first to avoid the root, because the question asks for \( \lambda^2 \).

Answer: (b) \( 6 \)

Explanation:
1. Perpendicular means \( a+b+c=0 \).
2. Counting all triples from \( \{-2,-1,0,1,2\} \) with zero sum (checked by listing): 19 triples, including \( (0,0,0) \).
3. Non-zero vectors: \( n=18 \), so \( \dfrac n3=6 \). Answer: option (b).

Teacher's Note:
1. Do not forget to remove the zero vector from the count.
2. Organise the count by the value of one variable to avoid missing triples.

Answer: (a) \( 1 \)

Explanation:
1. \( \vec d \) is along \( \vec a+\vec b+\vec c \), whose square of magnitude is \( 3+6\cos\theta \).
2. \( \cos\alpha=\dfrac{\vec d\cdot\vec a}{1}=\dfrac{1+2\cos\theta}{\sqrt{3+6\cos\theta}} \), so \( \cos^2\alpha=\dfrac{1+2\cos\theta}{3} \).
3. \( \cos2\alpha=2\cos^2\alpha-1=\dfrac{4\cos\theta-1}{3} \).
4. \( 4\cos\theta-3\cos2\alpha=4\cos\theta-(4\cos\theta-1)=1 \). Answer: option (a).

Teacher's Note:
1. A vector equally inclined to three unit vectors points along their sum.
2. Use the double-angle form \( \cos2\alpha=2\cos^2\alpha-1 \) to avoid inverse functions.

Answer: (b) \( 5 \)

Explanation:
1. Take the right angle at the origin and the acute-angle vertices at \( (1,0) \) and \( (0,1) \).
2. The medians go from \( (1,0) \) to \( (0,\tfrac12) \) and from \( (0,1) \) to \( (\tfrac12,0) \), with directions \( (-1,\tfrac12) \) and \( (\tfrac12,-1) \).
3. \( \cos\theta=\dfrac{|-\tfrac12-\tfrac12|}{\tfrac54}=\dfrac45 \), so \( \sec\theta=\dfrac54 \) and \( 4\sec\theta=5 \). Answer: option (b).

Teacher's Note:
1. Place the triangle on coordinate axes to turn the geometry into a dot-product calculation.
2. Take the absolute value of the dot product to get the acute angle.

Answer: (d) \( 3 \)

Explanation:
1. Check the direction: \( (1,3,-4)\cdot\vec V_3=-2+6-4=0 \), and the determinant of \( \vec V_1,\vec V_2,(1,3,-4) \) is 0, so it lies in their plane.
2. Projection of \( \vec V \) on \( \hat i-\hat j+\hat k \): \( \dfrac{t(1-3-4)}{\sqrt3}=-2\sqrt3\,t \).
3. \( |{-2\sqrt3\,t}|=6\sqrt3 \) gives \( |t|=3 \). Answer: option (d).

Teacher's Note:
1. Verify the given direction against both conditions before using the projection.
2. Magnitude of the projection means the sign of \( t \) does not matter.

Answer: (c) \( 2 \)

Explanation:
1. \( \hat n\times(\vec a\times\hat n)=\vec a-(\vec a\cdot\hat n)\hat n \) is the component of \( \vec a \) parallel to the plane.
2. The projection on the plane is \( \vec a-(\vec a\cdot\hat n-2)\hat n=\{\vec a-(\vec a\cdot\hat n)\hat n\}+2\hat n \).
3. So \( \vec b=2\hat n \) and \( |\vec b|=2 \), independent of \( |\vec a| \). Answer: option (c).

Teacher's Note:
1. Split the vector into a part parallel to the plane and a part along the normal.
2. The length of the normal part is fixed by the plane equation.

Answer: (b) \( 2 \)

Explanation:
1. Use \( (\vec p\times\vec q)\times\vec s=(\vec p\cdot\vec s)\vec q-(\vec q\cdot\vec s)\vec p \) on each term and combine.
2. Each term gives \( [\vec a\,\vec b\,\vec c]\,\vec r \) minus a part that cancels in the sum, and the sum equals \( 2[\vec a\,\vec b\,\vec c]\vec r \) (also checked numerically with random vectors).
3. Hence \( \lambda=2 \). Answer: option (b).

Teacher's Note:
1. Take \( \vec a\times\vec b \) as one vector and expand the triple cross product.
2. Testing with random numbers is a quick confirmation for identities like this.

Answer: (c) \( 6 \)

Explanation:
1. A point of line \( AB \): \( (1+s,\,2-s,\,1+s) \).
2. Normal to plane \( CDE \): \( \overrightarrow{CD}\times\overrightarrow{CE}=(2,2,-2)\times(4,5,-2)=(6,-4,2)\) up to scale; the plane is \( 3x-2y+z=k \), and \( C(0,-4,4) \) gives \( k=12 \).
3. Substitute: \( 3(1+s)-2(2-s)+(1+s)=12 \), i.e. \( 6s=12 \), \( s=2 \).
4. \( P=(3,0,3) \), so \( \alpha+\beta+\gamma=6 \). Answer: option (c).

Teacher's Note:
1. Write the plane through three points using the cross product of two edge vectors.
2. Put the parametric point of the line into the plane equation to find the parameter.

Answer: (c) \( 1 \)

Explanation:
1. Coplanar vectors have a zero Gram determinant: \( \begin{vmatrix}1&\cos\gamma&\cos\beta\\\cos\gamma&1&\cos\alpha\\\cos\beta&\cos\alpha&1\end{vmatrix}=0 \).
2. Expanding: \( 1-\cos^2\alpha-\cos^2\beta-\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=0 \).
3. So the required expression equals 1. Answer: option (c).

Teacher's Note:
1. Three vectors are coplanar exactly when the determinant of their pairwise dot products is zero.
2. Expand the determinant and rearrange to match the required expression.

Answer: (d) \( 3 \)

Explanation:
1. Coplanar: \( \begin{vmatrix}-1&c&b\\-c&1&a\\b&a&-1\end{vmatrix}=1+a^2-b^2-c^2=0 \) (expanded with sympy), so \( b^2+c^2=1+a^2 \).
2. \( |a\hat i+b\hat j+c\hat k|^2=a^2+b^2+c^2=1+2a^2 \).
3. With \( |a|\le1 \) the maximum is \( 1+2=3 \). Answer: option (d).

Teacher's Note:
1. Coplanarity gives a relation between the unknowns; substitute it into the expression.
2. Then apply the bound on the one variable that remains.

Answer: (c) \( 6 \)

Explanation:
1. Let the edges be \( \vec{a} \) and \( \vec{b} \) along the two opposite edges. Volume \( =\tfrac16|[\vec a\ \vec b\ \overrightarrow{PQ}]| \) where \( PQ \) joins the edges.
2. \( |\vec a\times\vec b|=ab\sin\theta \) and the component of \( \overrightarrow{PQ} \) along \( \vec a\times\vec b \) is the shortest distance \( d \).
3. Volume \( =\dfrac16abd\sin\theta \), so \( k=6 \). Answer: option (c).

Teacher's Note:
1. The tetrahedron volume is one sixth of the scalar triple product of three edges from a vertex.
2. The shortest distance is the projection on the common perpendicular.

Answer: (b) \( 2 \)

Explanation:
1. Take the cube with vertices at \( (0,0,0) \) to \( (1,1,1) \); the diagonal through the origin has direction \( (1,1,1) \). The corner \( (1,0,0) \) is not on it.
2. Distance \( =\dfrac{|(1,0,0)\times(1,1,1)|}{\sqrt3}=\dfrac{|(0,-1,1)|}{\sqrt3}=\sqrt{\dfrac23} \).
3. \( 3k^2=3\cdot\dfrac23=2 \). Answer: option (b).

Teacher's Note:
1. Place the cube with a vertex at the origin so that diagonals have simple directions.
2. Use distance from a point to a line with a cross product.

Answer: (c) \( 1 \)

Explanation:
1. The line has direction \( (4,0,-3) \) and the \( xy \)-plane has normal \( \hat k \), so \( \sin\alpha=\dfrac{3}{5} \) and \( \cos^2\alpha=\dfrac{16}{25} \).
2. The planes have normals \( (1,2,0) \) and \( (2,1,0) \): \( \cos\beta=\dfrac{4}{5} \), so \( \sin^2\beta=\dfrac{9}{25} \).
3. \( \cos^2\alpha+\sin^2\beta=\dfrac{16+9}{25}=1 \). Answer: option (c).

Teacher's Note:
1. The angle between a line and a plane uses the sine with the normal, while the angle between planes uses the cosine of the normals.
2. Square before adding to keep values simple.