JEE Mathematics Vector Algebra MCQs Set 28

Answer: (d) exactly one value of \( (p,q) \)

Explanation:
1. \( [3\vec u\ p\vec v\ p\vec w]=3p^2[\vec u\ \vec v\ \vec w] \), \( [p\vec v\ \vec w\ q\vec u]=pq[\vec u\ \vec v\ \vec w] \) and \( [2\vec w\ q\vec v\ q\vec u]=-2q^2[\vec u\ \vec v\ \vec w] \).
2. The equation becomes \( (3p^2-pq+2q^2)[\vec u\ \vec v\ \vec w]=0 \), and \( [\vec u\ \vec v\ \vec w]\ne0 \), so \( 3p^2-pq+2q^2=0 \).
3. This quadratic form has discriminant \( q^2-24q^2\lt0 \) unless \( q=0 \), so the only solution is \( p=q=0 \). Answer: (d)

Teacher's Note:
1. Expand each scalar triple product in terms of \( [\vec u\ \vec v\ \vec w] \).
2. Check when the resulting quadratic form vanishes.

Answer: (a) \( \pi \)

Explanation:
1. \( \vec a=8\vec b \) is in the same direction as \( \vec b \), and \( \vec c=-7\vec b \) is in the opposite direction.
2. So \( \vec a \) and \( \vec c \) are anti-parallel and the angle between them is \( \pi \). Answer: (a)

Teacher's Note:
1. Compare the directions of \( \vec a \) and \( \vec c \) with \( \vec b \).
2. Parallel and anti-parallel vectors have angle 0 and \( \pi \).

Answer: (d) -2

Explanation:
1. \( \vec a,\vec b,\vec c \) are coplanar, so \( \begin{vmatrix}1&1&1\\1&-1&2\\x&x-2&-1\end{vmatrix}=0 \).
2. Expanding: \( 1(1-2(x-2))-1(-1-2x)+1(x-2+x)=4+2x=0 \), so \( x=-2 \). Answer: (d)

Teacher's Note:
1. Use the scalar triple product.
2. Expand the determinant.

Answer: (c) parallel

Explanation:
1. Expand both sides: \( (\vec a\cdot\vec c)\vec b-(\vec b\cdot\vec c)\vec a=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c \).
2. So \( (\vec b\cdot\vec c)\vec a=(\vec a\cdot\vec b)\vec c \), i.e. \( \vec a=\frac{\vec a\cdot\vec b}{\vec b\cdot\vec c}\vec c \) (the denominator is non-zero). Hence \( \vec a \) is parallel to \( \vec c \). Answer: (c)

Teacher's Note:
1. Expand both vector triple products.
2. Cancel the common term.

Answer: (d) \( \overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC} \)

Explanation:
1. Taking \( P \) as the origin, the position vector of the mid-point is the average: \( \overrightarrow{PC}=\frac{\overrightarrow{PA}+\overrightarrow{PB}}2 \).
2. So \( \overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC} \). Answer: (d)

Teacher's Note:
1. Use the mid-point formula with \( P \) as origin.

Answer: (b) \( 2\vec a^2 \)

Explanation:
1. Let \( \vec a=a_1\hat i+a_2\hat j+a_3\hat k \). Then \( \vec a\times\hat i=a_3\hat j-a_2\hat k \), \( \vec a\times\hat j=a_1\hat k-a_3\hat i \), \( \vec a\times\hat k=a_2\hat i-a_1\hat j \).
2. The squares are \( a_2^2+a_3^2 \), \( a_1^2+a_3^2 \) and \( a_1^2+a_2^2 \), whose sum is \( 2(a_1^2+a_2^2+a_3^2)=2\vec a^2 \). Answer: (b)

Teacher's Note:
1. Write \( \vec a \) in components.
2. Square the magnitudes of the three cross products.

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

Explanation:
1. \( [\vec a\ \vec b\ \vec c]=\begin{vmatrix}1&0&-1\\x&1&1-x\\y&x&1+x-y\end{vmatrix} \).
2. Apply \( C_3\to C_3+C_1 \): the third column becomes \( (0,1,1+x) \). Expanding along the first row, the determinant is \( 1\cdot(1\cdot(1+x)-x)=1 \).
3. The value is 1, independent of \( x \) and \( y \). Answer: (a)

Teacher's Note:
1. Write the scalar triple product as a determinant.
2. Simplify with a column operation.

Answer: (c) no value of \( \lambda \)

Explanation:
1. LHS \( =\lambda\cdot\lambda^2\cdot\lambda[\vec a+\vec b\ \ \vec b\ \ \vec c]=\lambda^4[\vec a\ \vec b\ \vec c] \).
2. RHS \( =[\vec a\ \ \vec c\ \ \vec b]=-[\vec a\ \vec b\ \vec c] \) (the term with \( \vec b,\vec b \) vanishes).
3. So \( \lambda^4=-1 \), which has no real solution. Answer: (c)

Teacher's Note:
1. Pull scalar factors out of the triple product.
2. Use the antisymmetry of the triple product.

Answer: (a) 40 units

Explanation:
1. The total force is \( \vec F=7\hat i+2\hat j-4\hat k \) and the displacement is \( \vec d=4\hat i+2\hat j-2\hat k \).
2. Work done \( =\vec F\cdot\vec d=28+4+8=40 \) units. Answer: (a)

Teacher's Note:
1. Add the forces.
2. Take the dot product with the displacement.

Answer: (c) all except two values of \( \lambda \)

Explanation:
1. The three vectors are non-coplanar if \( \begin{vmatrix}1&2&3\\0&\lambda&4\\0&0&2\lambda-1\end{vmatrix}\ne0 \), and this determinant (upper triangular) is \( \lambda(2\lambda-1) \).
2. It is non-zero unless \( \lambda=0 \) or \( \lambda=\frac12 \). So the vectors are non-coplanar for all values except two. Answer: (c)

Teacher's Note:
1. Write the determinant of coefficients.
2. Find where it vanishes.

Answer: (a) \( \cos^{-1}\left(\frac{19}{35}\right) \)

Explanation:
1. Normal to \( OAB \): \( \vec n_1=\overrightarrow{OA}\times\overrightarrow{OB}=5\hat i-\hat j-3\hat k \). Normal to \( ABC \): \( \vec n_2=\overrightarrow{AB}\times\overrightarrow{AC}=\hat i-5\hat j-3\hat k \).
2. \( \cos\theta=\frac{\vec n_1\cdot\vec n_2}{|\vec n_1||\vec n_2|}=\frac{5+5+9}{\sqrt{35}\sqrt{35}}=\frac{19}{35} \). Answer: (a)

Teacher's Note:
1. Find the normals to the two faces.
2. Use the angle between normals.

Answer: (b) \( \vec u\cdot\vec v\times\vec w \)

Explanation:
1. \( (\vec u-\vec v)\times(\vec v-\vec w)=\vec u\times\vec v-\vec u\times\vec w-\vec v\times\vec v+\vec v\times\vec w=\vec u\times\vec v-\vec u\times\vec w+\vec v\times\vec w \).
2. Dot with \( \vec u+\vec v-\vec w \). Triple products with a repeated vector vanish, so only three terms survive: \( \vec u\cdot(\vec v\times\vec w)=[\vec u\ \vec v\ \vec w] \), \( \vec v\cdot(-\vec u\times\vec w)=[\vec u\ \vec v\ \vec w] \) and \( -\vec w\cdot(\vec u\times\vec v)=-[\vec u\ \vec v\ \vec w] \).
3. The result is \( \vec u\cdot\vec v\times\vec w \). Answer: (b)

Teacher's Note:
1. Expand the cross product.
2. Drop triple products with a repeated vector.

Answer: (d) none of these

Explanation:
1. \( AB=\sqrt{36+4+9}=7 \), \( BC=\sqrt{4+9+36}=7 \), \( CD=\sqrt{36+4+1}=\sqrt{41} \) and \( DA=\sqrt{4+9+4}=\sqrt{17} \).
2. Opposite sides are not equal (\( AB\ne CD \)), so \( ABCD \) is not a parallelogram and hence not a square, rhombus or rectangle. Answer: (d)

Teacher's Note:
1. Find the four side lengths.
2. A rhombus, rectangle or square needs opposite sides equal.

Answer: (b) -1

Explanation:
1. Split the third column: the determinant \( =\begin{vmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{vmatrix}+abc\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix} \) (after taking \( a,b,c \) out of the rows of the second determinant).
2. Both determinants are equal in value (row/column permutations of the same Vandermonde determinant), so the expression is \( (1+abc)\times\Delta=0 \) with \( \Delta\ne0 \) by non-coplanarity.
3. Hence \( 1+abc=0 \), i.e. \( abc=-1 \). Answer: (b)

Teacher's Note:
1. Split the determinant by its last column.
2. Use non-coplanarity to say the common determinant is non-zero.

Answer: (d) 3

Explanation:
1. \( \hat n \) is perpendicular to both \( \vec u \) and \( \vec v \), so \( \hat n=\pm\frac{\vec u\times\vec v}{|\vec u\times\vec v|} \). Here \( \vec u\times\vec v=-2\hat k \), so \( \hat n=\pm\hat k \).
2. \( |\vec w\cdot\hat n|=|3|=3 \). Answer: (d)

Teacher's Note:
1. Find \( \hat n \) from the cross product.
2. Take the dot product with \( \vec w \).

Answer: (a) \( \frac1{\sqrt2}(\hat i+\hat j) \)

Explanation:
1. Both vectors lie in the XY-plane, so a coplanar vector has the form \( \vec a=x\hat i+y\hat j \).
2. \( \vec a\perp(\hat i-\hat j)\Rightarrow x-y=0\Rightarrow\vec a=x(\hat i+\hat j) \), with \( |\vec a|=|x|\sqrt2 \).
3. A unit vector is \( \frac1{\sqrt2}(\hat i+\hat j) \). Answer: (a)

Teacher's Note:
1. Write a general vector in the plane.
2. Impose perpendicularity and unit length.

Answer: (a) \( \left\{-\frac23,2\right\} \)

Explanation:
1. Doubling the magnitude: \( 2\sqrt{1+x^2+9}=\sqrt{16+(4x-2)^2+4} \).
2. Squaring: \( 4(10+x^2)=20+(4x-2)^2\Rightarrow40+4x^2=20+16x^2-16x+4\Rightarrow3x^2-4x-4=0 \).
3. \( (x-2)(3x+2)=0\Rightarrow x=2 \) or \( x=-\frac23 \). Answer: (a)

Teacher's Note:
1. Equate the magnitudes.
2. Solve the resulting quadratic.

Answer: (b) \( \vec a\times\vec b=\vec b\times\vec c=\vec c\times\vec a \)

Explanation:
1. The sides of a triangle taken in order add to zero: \( \vec a+\vec b+\vec c=\vec 0 \).
2. Then \( \vec a+\vec b=-\vec c \) gives \( (\vec a+\vec b)\times\vec b=-\vec c\times\vec b \), so \( \vec a\times\vec b=\vec b\times\vec c \). Similarly \( \vec b\times\vec c=\vec c\times\vec a \).
3. So \( \vec a\times\vec b=\vec b\times\vec c=\vec c\times\vec a \) (each equals twice the area vector). Answer: (b)

Teacher's Note:
1. Use \( \vec a+\vec b+\vec c=\vec 0 \).
2. Cross with one of the vectors.

Answer: (a) \( z\hat i-x\hat k \)

Explanation:
1. \( \vec a,\vec c,\vec b \) form a right handed system means \( \vec c=\vec b\times\vec a \) (up to the length, and the rule gives exactly \( \vec b\times\vec a \) in the standard cyclic order).
2. \( \vec b\times\vec a=\hat j\times(x\hat i+y\hat j+z\hat k)=-x\hat k+z\hat i=z\hat i-x\hat k \). Answer: (a)

Teacher's Note:
1. Use the cyclic order for a right handed system.
2. Compute \( \hat j\times\vec a \).