JEE Mathematics Vector Algebra MCQs Set 07

Answer: (b) \( \frac1{15}(−11\hat{i}−10\hat{j}−2\hat{k}) \)

Explanation:
1. The bisector of two vectors is along the sum of their unit vectors: \( \vec{c}+\frac{3\hat{i}+4\hat{j}}{5}=λ(−\hat{i}+\hat{j}−\hat{k}) \).
2. So \( \vec{c}=(−λ−\frac35,\ λ−\frac45,\ −λ) \).
3. \( |\vec{c}|=1 \) gives \( 3λ^2−\frac25λ=0 \); \( λ=0 \) would make \( \vec{c} \) opposite to the other unit vector (no bisector), so \( λ=\frac2{15} \).
4. Then \( \vec{c}=\frac1{15}(−11,−10,−2) \).

Teacher's Note:
1. The sum of two unit vectors lies along the angle bisector.
2. Reject the root that makes the sum zero.

Answer: (b) \( \frac32 \)

Explanation:
1. Take \( A \) as origin, \( \overrightarrow{AB}=\vec{b} \) and \( \overrightarrow{AD}=\vec{d} \). Then \( \overrightarrow{AL}=\vec{b}+\frac12\vec{d} \) and \( \overrightarrow{AM}=\vec{d}+\frac12\vec{b} \).
2. The sum is \( \frac32(\vec{b}+\vec{d})=\frac32\overrightarrow{AC} \).
3. Hence \( K=\frac32 \).

Teacher's Note:
1. Express every vector in terms of two adjacent sides.
2. The diagonal \( AC \) equals \( \vec{b}+\vec{d} \).

Answer: (b) \( (−∞,−6]∪\left[−\frac16,∞\right) \)

Explanation:
1. The initial point is \( \frac{k(5\hat{j})+5\hat{i}}{k+1}=\left(\frac5{k+1},\frac{5k}{k+1}\right) \), so \( |\vec{b}|^2=\frac{25(1+k^2)}{(k+1)^2}≤37 \).
2. This gives \( 6k^2+37k+6≥0 \), i.e. \( (6k+1)(k+6)≥0 \).
3. So \( k≤−6 \) or \( k≥−\frac16 \) (with \( k≠−1 \), and \( k=−1 \) is not in these ranges).

Teacher's Note:
1. Use the section formula and square the magnitude to remove the root.
2. Solve the quadratic inequality using its roots.

Answer: (d) \( \sqrt6 \)

Explanation:
1. Here \( \vec{a}−\vec{b}=(4,−2,−6)=\vec{c} \), so \( \vec{a}=\vec{b}+\vec{c} \): the sides \( \vec{b},\vec{c} \) are traversed one after the other and \( \vec{a} \) closes the triangle.
2. The median from the vertex opposite \( \vec{c} \) to the mid-point of \( \vec{c} \) is \( \frac12(\vec{a}+\vec{b})=\frac12(2,4,2)=(1,2,1) \).
3. Its length is \( \sqrt6 \).

Teacher's Note:
1. First find which side is the sum of the other two.
2. The median from a vertex is half the sum of the two sides meeting there.

Answer: (a) 1

Explanation:
1. A plane through \( P,Q,R,S \) gives, by the condition for four coplanar points (Menelaus' theorem for a skew quadrilateral), \( \frac{AP}{PB}·\frac{BQ}{QC}·\frac{CR}{RD}·\frac{DS}{SA}=1 \).
2. Hence the product is 1.

Teacher's Note:
1. This is Menelaus' theorem extended to four sides.
2. The ratios are taken as unsigned lengths.

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

Explanation:
1. \( \overrightarrow{BC}=(−2,−6,3) \) with \( |\overrightarrow{BC}|=7 \), and \( \overrightarrow{BA}=(−1,3,0) \).
2. \( \overrightarrow{BA}×\overrightarrow{BC}=(9,3,12) \) with magnitude \( \sqrt{234}=3\sqrt{26} \).
3. Distance \( =\dfrac{|\overrightarrow{BA}×\overrightarrow{BC}|}{|\overrightarrow{BC}|}=\dfrac{3\sqrt{26}}7 \).

Teacher's Note:
1. Distance of a point from a line is area of the parallelogram divided by the base.
2. Compute the cross product carefully.

Answer: (a) Parallelogram, which is neither a rhombus nor a rectangle

Explanation:
1. \( \overrightarrow{PQ}=(6,1) \), \( \overrightarrow{QR}=(−1,3) \), \( \overrightarrow{SR}=(6,1) \).
2. \( \overrightarrow{PQ}=\overrightarrow{SR} \), so it is a parallelogram.
3. \( |PQ|=\sqrt{37}≠|QR|=\sqrt{10} \) (not a rhombus) and \( \overrightarrow{PQ}·\overrightarrow{QR}=−3≠0 \) (not a rectangle).

Teacher's Note:
1. Check opposite sides first, then adjacent sides and angles.
2. A parallelogram with unequal adjacent sides and no right angle is neither a rhombus nor a rectangle.

Answer: (b) \( \dfrac{\sqrt{17}}{9} \)

Explanation:
1. \( |\overrightarrow{AB}|=15 \), \( |\overrightarrow{AD}|=3 \) and \( \overrightarrow{AB}·\overrightarrow{AD}=−2+20+22=40 \).
2. The angle \( θ \) between them has \( \cosθ=\frac{40}{45}=\frac89 \) and \( \sinθ=\frac{\sqrt{17}}9 \).
3. After the rotation the angle becomes \( 90^\circ \), so \( α=90^\circ−θ \) and \( \cosα=\sinθ=\frac{\sqrt{17}}9 \).

Teacher's Note:
1. Find the angle between the two sides and see how much extra rotation gives 90 degrees.
2. The cosine of the complementary angle is the sine.

Answer: (b) \( \vec{a}+\vec{b} \)

Explanation:
1. \( \vec{r}×\vec{a}=\vec{b}×\vec{a} \) gives \( (\vec{r}−\vec{b})×\vec{a}=\vec0 \), so \( \vec{r}=\vec{b}+λ\vec{a} \).
2. \( \vec{r}×\vec{b}=\vec{a}×\vec{b} \) gives \( (\vec{r}−\vec{a})×\vec{b}=\vec0 \), so \( \vec{r}=\vec{a}+μ\vec{b} \).
3. Equating, \( λ=μ=1 \) and \( \vec{r}=\vec{a}+\vec{b} \).

Teacher's Note:
1. A cross-product equation of this type means that the difference of two vectors is parallel to a third.
2. Compare the coefficients of \( \vec{a} \) and \( \vec{b} \).

Answer: (a) \( \sqrt{\dfrac23}\,a \)

Explanation:
1. The centroid of the base (an equilateral triangle of side \( a \)) is at distance \( \frac{a}{\sqrt3} \) from each base vertex.
2. Height \( h=\sqrt{a^2−\frac{a^2}3} \).
3. So \( h=\sqrt{\frac23}\,a \).

Teacher's Note:
1. The foot of the perpendicular is the centroid of the opposite face.
2. Use Pythagoras in the right triangle formed by the edge, the height and the centroid distance.

Answer: (c) \( \dfrac{k}{\sqrt2} \)

Explanation:
1. Place the tetrahedron at alternate vertices of a cube of side \( s \), so that the edge is \( k=s\sqrt2 \).
2. Two opposite edges lie on opposite faces of the cube, and are perpendicular, so the distance between them is the cube side \( s \).
3. \( s=\frac{k}{\sqrt2} \).

Teacher's Note:
1. A regular tetrahedron can be inscribed in a cube.
2. Opposite edges are diagonals of opposite faces.

Answer: (c) \( 4\sqrt3 \)

Explanation:
1. \( \vec{d}_1×\vec{d}_2=(2,−7,5)×(2,1,−3)=(16,16,16) \) with magnitude \( 16\sqrt3 \).
2. The vector joining points on the lines is \( (−1−3,\ 1+15,\ 9−9)=(−4,16,0) \).
3. Distance \( =\frac{|(−4,16,0)·(16,16,16)|}{16\sqrt3}=\frac{192}{16\sqrt3}=4\sqrt3 \).

Teacher's Note:
1. Use the formula distance = scalar triple product over magnitude of the cross product.
2. Choose any one point from each line.

Answer: (a) \( \cos^{-1}\left(\dfrac{3}{\sqrt{30}}\right) \)

Explanation:
1. Normals: \( (1,1,1)×(2,0,0)=(0,2,−2)∥(0,1,−1) \) and \( (1,1,0)×(0,1,−1)=(−1,1,1) \).
2. Line of intersection: \( (0,1,−1)×(−1,1,1)=(2,1,1) \).
3. \( \cosθ=\frac{|(2,1,1)·(2,−1,0)|}{\sqrt6\sqrt5}=\frac{3}{\sqrt{30}} \).

Teacher's Note:
1. The line of intersection is along the cross product of the two normals.
2. The angle between a line and a vector uses the dot product of directions.

Answer: (c) \( \left|\dfrac{q−\vec{a}·\vec{n}}{|\vec{n}|}\right| \)

Explanation:
1. The unit normal is \( \vec{n}/|\vec{n}| \) and the plane is at signed distance \( q/|\vec{n}| \) from the origin.
2. The point is at signed distance \( \vec{a}·\vec{n}/|\vec{n}| \) along the normal.
3. The difference is \( \left|\frac{q−\vec{a}·\vec{n}}{|\vec{n}|}\right| \).

Teacher's Note:
1. Compare the projection of the point on the normal with that of the plane.
2. Take the modulus for distance.

Answer: (a) \( \vec{a}+2\left(\dfrac{q−\vec{a}·\vec{n}}{|\vec{n}|^2}\right)\vec{n} \)

Explanation:
1. The foot of the perpendicular is \( \vec{a}+λ\vec{n} \) with \( (\vec{a}+λ\vec{n})·\vec{n}=q \), so \( λ=\frac{q−\vec{a}·\vec{n}}{|\vec{n}|^2} \).
2. The reflection is twice as far from \( \vec{a} \) along the normal: \( \vec{a}+2λ\vec{n} \).

Teacher's Note:
1. First find the foot of the perpendicular; the image is as far again.
2. The reflection is not the foot.

Answer: (a) \( \left|\dfrac{q−\vec{a}·\vec{n}}{\vec{c}·\vec{n}}\right|\,|\vec{c}| \)

Explanation:
1. Move from \( \vec{a} \) along \( \vec{c} \): the point \( \vec{a}+λ\vec{c} \) lies on the plane when \( (\vec{a}+λ\vec{c})·\vec{n}=q \), so \( λ=\frac{q−\vec{a}·\vec{n}}{\vec{c}·\vec{n}} \).
2. The distance travelled is \( |λ||\vec{c}| \).

Teacher's Note:
1. Parametrise the line through the point with the given direction.
2. Distance equals the parameter multiplied by the length of the direction vector.

Answer: (b) \( \vec{r}=\vec{a}+t(\vec{n}_1×\vec{n}_2) \)

Explanation:
1. The line of intersection lies in both planes, so it is perpendicular to both normals.
2. Its direction is therefore \( \vec{n}_1×\vec{n}_2 \).
3. The required line is \( \vec{r}=\vec{a}+t(\vec{n}_1×\vec{n}_2) \).

Teacher's Note:
1. A line lying in a plane is perpendicular to the plane's normal.
2. The common perpendicular direction of two vectors is their cross product.

Answer: (b) \( \vec0 \)

Explanation:
1. For three planes meeting in a point, \( \vec{r}[\vec{n}_1\ \vec{n}_2\ \vec{n}_3]=p_1(\vec{n}_2×\vec{n}_3)+p_2(\vec{n}_3×\vec{n}_1)+p_3(\vec{n}_1×\vec{n}_2) \).
2. When they share a common line, the normals are coplanar and \( [\vec{n}_1\ \vec{n}_2\ \vec{n}_3]=0 \).
3. So the expression is the null vector.

Teacher's Note:
1. Recall the vector form of the solution of three plane equations.
2. A common line implies coplanar normals.

Answer: (a) \( V≤L^3 \)

Explanation:
1. For non-negative entries, \( |\det|≤(a_1+a_2+a_3)(b_1+b_2+b_3)(c_1+c_2+c_3) \) (each term of the determinant expansion appears in the product with a positive sign).
2. The three factors add to \( 3L \), so by AM-GM their product is at most \( L^3 \).
3. Hence \( V≤L^3 \).

Teacher's Note:
1. Bound the determinant by the product of the row sums.
2. Then apply the AM-GM inequality to the three row sums.

Answer: (c) \( \dfrac{2\pi}3 \)

Explanation:
1. The scalar triple product squared equals the Gram determinant: \( [\vec{a}\ \vec{b}\ \vec{c}]^2=\begin{vmatrix}1&c&c\\c&1&c\\c&c&1\end{vmatrix}=1−3c^2+2c^3 \), with \( c=\cosθ \).
2. This must be non-negative: \( (1−c)^2(1+2c)≥0 \), so \( c≥−\frac12 \).
3. Thus \( θ≤\frac{2\pi}3 \).

Teacher's Note:
1. The square of a real number cannot be negative.
2. Factor the Gram determinant to see the allowed range of the cosine.

Answer: (a) \( \dfrac1{\sqrt2} \) cubic unit

Explanation:
1. \( V^2=[\hat a\ \hat b\ \hat c]^2=\begin{vmatrix}1&\frac12&\frac12\\\frac12&1&\frac12\\\frac12&\frac12&1\end{vmatrix} \).
2. The determinant is \( 1−3\cdot\frac14+2\cdot\frac18=\frac12 \).
3. So \( V=\frac1{\sqrt2} \) cubic unit.

Teacher's Note:
1. The square of the scalar triple product equals the Gram determinant of dot products.
2. Use the shortcut \( 1−3c^2+2c^3 \) for equal dot products.

Answer: (c) \( 3\hat{i}−\hat{j}+3\hat{k} \)

Explanation:
1. Write \( \vec{v}=α\vec{a}+β\vec{b}=(α+β,\ α−β,\ α+β) \).
2. The projection on \( \vec{c} \) (with \( \vec{c}·\vec{a}=−1, \vec{c}·\vec{b}=1, |\vec{c}|=\sqrt3 \)) is \( \frac{−α+β}{\sqrt3}=\frac1{\sqrt3} \), so \( β−α=1 \).
3. Only \( 3\hat{i}−\hat{j}+3\hat{k} \) is of the form \( (x,y,x) \) with \( x=α+β=3, y=α−β=−1 \) (so \( α=1, β=2 \), \( β−α=1 \)). The other options either have unequal \( i \) and \( k \) components or do not satisfy \( β−α=1 \).

Teacher's Note:
1. A vector in the plane of \( \vec{a} \) and \( \vec{b} \) is a combination of them.
2. Test the options by the form \( (x,y,x) \) and the projection condition.

Answer: (c) All \( x<0 \)

Explanation:
1. Acute angle: \( \vec{a}·\vec{b}=2x^2−3x+1>0 \), i.e. \( (2x−1)(x−1)>0 \), so \( x<\frac12 \) or \( x>1 \).
2. Obtuse angle with the y-axis: \( \vec{b}·\hat{j}=x<0 \).
3. Both hold exactly for \( x<0 \).

Teacher's Note:
1. Acute angle means positive dot product, obtuse means negative.
2. Intersect the two conditions.

Answer: (a) All values of \( x \)

Explanation:
1. Subtract the first row from the second and the second from the third: both rows become \( (3,3,3) \).
2. Two equal rows make the determinant zero for every \( x \).
3. So the vectors are coplanar for all \( x \).

Teacher's Note:
1. Row operations do not change a determinant.
2. Equal rows make it vanish.

Answer: (a) (i) and (ii) only

Explanation:
1. \( |\vec{a}|^2=2 \) and \( \vec{a}·\vec{b}=1 \), so the common angle has \( \cos=\frac12 \). For \( \vec{c}=(x,y,z) \): \( x+y=1 \), \( y+z=1 \) and \( x^2+y^2+z^2=2 \).
2. So \( x=z=1−y \) and \( 3y^2−4y=0 \), giving \( y=0 \) or \( y=\frac43 \).
3. The solutions are \( \vec{c}=(1,0,1)=\hat{i}+\hat{k} \) and \( \vec{c}=(−\frac13,\frac43,−\frac13)=−\frac13(\hat{i}−4\hat{j}+\hat{k}) \), i.e. (i) and (ii).

Teacher's Note:
1. Write the three conditions (equal length, two equal dot products) as equations in the components.
2. A quadratic can give two valid vectors.

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. With centre \( O=(6,−1) \), \( \overrightarrow{OA}=(−5,−1) \). Rotating anticlockwise by 90° gives \( \overrightarrow{OB}=(1,−5) \), so \( B=(7,−6) \).
2. \( D=2O−B=(5,4) \) and \( C=2O−A=(11,0) \) (not \( (10,1) \)).
3. \( AC=(10,2) \), so \( AC^2=104 \) and the area is \( \frac12AC^2=52 \).

Teacher's Note:
1. Rotate the vector from the centre to one vertex by 90 degrees to get the next vertex.
2. The area of a square equals half the square of its diagonal.

Answer: (d) All four

Explanation:
1. Dotting with \( \vec{a} \) and \( \vec{b} \) gives \( α=\vec{c}·\vec{a}=\cosθ \) and \( β=\vec{c}·\vec{b}=\cosθ \), so (i) holds.
2. \( |\vec{c}|^2=1 \) gives \( α^2+β^2+γ^2=1 \), i.e. \( γ^2=1−2α^2 \), which is (ii).
3. Then \( γ^2=1−2\cos^2θ=−\cos2θ \) (iii), and \( β^2=\cos^2θ=\frac{1+\cos2θ}2 \) (iv).

Teacher's Note:
1. Use dot products with \( \vec{a} \) and \( \vec{b} \) to find the coefficients.
2. Use the double angle formula for the cosine.