JEE Mathematics Vector Algebra MCQs Set 18

Answer: (b) a unit vector

Explanation:
1. Magnitude squared: \( \cos^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta+\sin^2\alpha=\cos^2\alpha+\sin^2\alpha=1 \).
2. The magnitude is 1, so it is a unit vector for every α and β (it is not fixed in direction, so it cannot be parallel to a given vector).

Teacher's Note:
1. The three coefficients are direction cosines of some line, which always satisfy l² + m² + n² = 1.
2. Check the magnitude first when a vector is given with trigonometric components.

Answer: (b) \( 5\hat i-11\hat j \)

Explanation:
1. Go from A to B, then B to C, then C to D: \( \overrightarrow{AD}=\overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CD} \).
2. \( =(2-1+4)\hat i+(-3-1-7)\hat j+(1-1+0)\hat k=5\hat i-11\hat j \).

Teacher's Note:
1. Reverse a vector when you traverse it backwards: BC = −CB.
2. Add the components of i, j, k separately.

Answer: (d) \( -\hat j+2\hat k,\ \hat i+\hat j+3\hat k \)

Explanation:
1. With A as origin, each position vector becomes (old vector) − (vector of A).
2. \( B:\ (\hat i+\hat j+\hat k)-(\hat i+2\hat j-\hat k)=-\hat j+2\hat k \).
3. \( C:\ (2\hat i+3\hat j+2\hat k)-(\hat i+2\hat j-\hat k)=\hat i+\hat j+3\hat k \).

Teacher's Note:
1. Shifting the origin to A subtracts the position vector of A from every point.
2. This is the same as computing AB and AC.

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

Explanation:
1. \( |\vec a|=\sqrt{36+9+4}=7 \).
2. The unit vector opposite to a is \( -\frac{\vec a}{|\vec a|}=\frac17(6\hat i-3\hat j+2\hat k) \).

Teacher's Note:
1. Opposite direction means multiply the unit vector by −1.
2. Divide by the magnitude, not its square.

Answer: (b) \( \frac13(2\hat i-\hat j+2\hat k) \)

Explanation:
1. \( \vec a+\vec b-\vec c=(2+1-5)\hat i+(1-1+1)\hat j+(-1+0-1)\hat k=-2\hat i+\hat j-2\hat k \).
2. \( \text{Magnitude}=\sqrt{4+1+4}=3 \).
3. The opposite unit vector is \( \frac13(2\hat i-\hat j+2\hat k) \).

Teacher's Note:
1. First find the resultant, then normalise it and reverse the sign.
2. Check that the magnitude of the final answer is 1.

Answer: (d) 3

Explanation:
1. Collinear vectors have proportional components: \( \frac{x}{-2}=\frac{-6}{3}=\frac{2}{y} \).
2. \( \frac{-6}{3}=-2\Rightarrow x=4,\ y=-1 \).
3. So x + y = 3.

Teacher's Note:
1. Set every ratio of corresponding components equal to a single constant.
2. The known ratio −6/3 = −2 fixes the other two.

Answer: (b) \( \frac32 \)

Explanation:
1. \( \overrightarrow{PQ}=3\hat i+3\sqrt3\hat j \), so \( |PQ|=\sqrt{9+27}=6 \).
2. \( \overrightarrow{RS}=2\hat i+2\sqrt3\hat j \), so \( |RS|=\sqrt{4+12}=4 \).
3. The ratio is 6/4 = 3/2.

Teacher's Note:
1. PQ = position vector of Q − position vector of P.
2. Note that PQ is parallel to RS here, with ratio 3 : 2.

Answer: (b) \( \pm3(2\hat i+3\hat j+6\hat k) \)

Explanation:
1. \( |\vec a|=\sqrt{4+9+36}=7 \).
2. A collinear vector of magnitude 21 is \( \pm21\cdot\frac{\vec a}{7}=\pm3(2\hat i+3\hat j+6\hat k) \).

Teacher's Note:
1. Collinear includes both the same and the opposite direction, hence the ± sign.
2. Magnitude of a vector = scalar × magnitude of the original.

Answer: (c) \( x=-\frac53,\ y=\frac65 \)

Explanation:
1. \( \frac x1=\frac{-2}{y}=\frac{5}{-3} \).
2. \( x=-\frac53 \) and \( y=\frac{-2\times(-3)}{5}=\frac65 \).

Teacher's Note:
1. Equate the three component ratios.
2. Be careful with the signs of the third ratio 5 / (−3).

Answer: (b) a = 1

Explanation:
1. Equality in the triangle inequality holds only when the two vectors have the same direction.
2. So p is a positive multiple of q, giving a = 1.

Teacher's Note:
1. |p + q| = |p| + |q| means the vectors are parallel and like.
2. Unlike (anti-parallel) vectors would give the minimum, not the maximum.

Answer: (a) 0

Explanation:
1. Coplanar: \( \begin{vmatrix}x&x-2&1\\1&1&1\\1&-1&2\end{vmatrix}=0 \).
2. \( x(2+1)-(x-2)(2-1)+1(-1-1)=3x-x+2-2=2x \).
3. So x = 0.

Teacher's Note:
1. A vector lies in the plane of two others when the determinant of the three is zero.
2. Expand along the first row carefully.

Answer: (d) right angled but not isosceles

Explanation:
1. Let A, B, C be the points in order. \( AB^2=1+4+36=41 \), \( BC^2=4+1+1=6 \), \( CA^2=1+9+25=35 \).
2. Since 41 = 6 + 35, the triangle is right angled at C.
3. The sides squared 41, 6, 35 are all different, so it is not isosceles.

Teacher's Note:
1. Compute the squared lengths of all three sides and test Pythagoras.
2. Equal squared lengths would be needed for an isosceles triangle.

Answer: (b) \( 3\vec a \)

Explanation:
1. In an equilateral triangle the orthocentre is also the centroid, so \( \vec a+\vec b+\vec c=\vec 0 \).
2. \( (\vec b-\vec a)+2(\vec c-\vec b)+3(\vec a-\vec c)=2\vec a-\vec b-\vec c \).
3. Using \( \vec b+\vec c=-\vec a \), this is \( 3\vec a \).

Teacher's Note:
1. For an equilateral triangle the orthocentre, centroid, circumcentre and incentre coincide.
2. Then the position vectors from this point add up to zero.

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

Explanation:
1. Rotation keeps the length, doubling doubles it: \( 16+(4x-2)^2+4=4(1+x^2+9) \).
2. So \( 16x^2-16x+24=4x^2+40\Rightarrow 12x^2-16x-16=0\Rightarrow 3x^2-4x-4=0 \).
3. \( (3x+2)(x-2)=0\Rightarrow x=-\frac23,\ 2 \).

Teacher's Note:
1. Only the magnitude is affected by the doubling; the rotation changes direction but not length.
2. Square both magnitudes and equate.

Answer: (b) \( \frac{11}{\sqrt{162}} \)

Explanation:
1. \( \overrightarrow{PQ}=4\hat i-5\hat j+11\hat k \), \( |PQ|=\sqrt{16+25+121}=\sqrt{162} \).
2. The cosine of the angle with the z-axis is the k-component divided by the magnitude: \( \frac{11}{\sqrt{162}} \).

Teacher's Note:
1. The angle with an axis uses the direction cosine: component along that axis divided by magnitude.
2. PQ is Q − P, not P − Q.

Answer: (b) \( \vec a=-4\vec b \)

Explanation:
1. \( |\vec b|=\sqrt{36+64+\frac{225}{4}}=\frac{25}{2} \).
2. So \( |\vec a|=50=4|\vec b| \), and \( \vec a=\pm4\vec b \).
3. The k-component of b is negative, and a must have a positive k-component (acute angle with the z-axis), so \( \vec a=-4\vec b \).

Teacher's Note:
1. Collinear and of magnitude 4|b| means a = ±4b.
2. The sign is decided by the sign of the z-component.

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

Explanation:
1. Taking P as the origin, \( \vec c=\frac{\vec a+\vec b}{2} \).
2. Hence \( \overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC} \).

Teacher's Note:
1. The position vector of the mid-point is the average of the end-point vectors, measured from any common origin.
2. P can be treated as the origin.

Answer: (d) 4 : 3

Explanation:
1. \( \vec c=\frac{3\vec a+4\vec b}{7} \).
2. Compare with \( \frac{m\vec b+n\vec a}{m+n} \): m = 4 and n = 3.
3. So AC : CB = 4 : 3.

Teacher's Note:
1. Because the coefficients 3 and 4 add up to 7, C lies on AB.
2. The coefficient of b gives m, the one of a gives n, and m : n = AC : CB.

Answer: (b) 1 : 2

Explanation:
1. Let the ratio be m : n. Using the x-coordinate: \( \frac{7m-2n}{m+n}=1\Rightarrow6m=3n\Rightarrow m:n=1:2 \).
2. Check with y: \( \frac{0\cdot m+3n}{m+n}=\frac{6}{3}=2 \) and with z: \( \frac{-m+5n}{m+n}=\frac93=3 \).

Teacher's Note:
1. Use one coordinate to find the ratio and the others to verify it.
2. A positive ratio means internal division.

Answer: (b) \( 2(\vec b+\vec c-2\vec a) \)

Explanation:
1. External division: \( \vec p=\frac{3\vec b-4\vec a}{3-4}=4\vec a-3\vec b \) and \( \vec q=\frac{2\vec c-\vec b}{2-1}=2\vec c-\vec b \).
2. \( \overrightarrow{PQ}=\vec q-\vec p=2\vec c-\vec b-4\vec a+3\vec b=2(\vec b+\vec c-2\vec a) \).

Teacher's Note:
1. For external division use (mb − na)/(m − n).
2. PQ is q − p.

Answer: (d) \( \hat i+\hat j+\hat k \)

Explanation:
1. All three direction cosines equal \( \frac1{\sqrt3} \).
2. So the direction is \( \frac{1}{\sqrt3}(\hat i+\hat j+\hat k) \), parallel to \( \hat i+\hat j+\hat k \).

Teacher's Note:
1. Equal angles with the three axes mean equal direction cosines.
2. The direction ratios are then 1 : 1 : 1.

Answer: (c) 1

Explanation:
1. \( l^2+m^2+n^2=1 \).
2. \( (l-m)^2+(m-n)^2+(n-l)^2\ge0\Rightarrow 2(l^2+m^2+n^2)-2(lm+mn+nl)\ge0 \).
3. So \( lm+mn+nl\le1 \), with equality when l = m = n.

Teacher's Note:
1. Use the identity (l − m)² + (m − n)² + (n − l)² ≥ 0.
2. Equality needs l = m = n = 1/√3.

Answer: (c) \( 2\vec a+3\vec b-\vec c=\vec0 \)

Explanation:
1. Write \( \vec c=x\vec a+y\vec b \): \( x+2y=8 \) and \( 2x+3y=13 \).
2. Solving: y = 3, x = 2. Check the k-component: 3x + y = 9. ✓
3. So \( 2\vec a+3\vec b-\vec c=\vec 0 \).

Teacher's Note:
1. Three coplanar vectors are linearly dependent; solve for two coefficients and check with the third equation.
2. Always verify with the unused component.

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

Explanation:
1. Let A(1,1,1), B(4,1,1), C(4,5,1). Then BC = 4, CA = 5, AB = 3 (a right angle at B).
2. Incentre \( =\frac{BC\cdot A+CA\cdot B+AB\cdot C}{12}=\frac{4(1,1,1)+5(4,1,1)+3(4,5,1)}{12}=(3,2,1) \).
3. So the incentre is \( 3\hat i+2\hat j+\hat k \).

Teacher's Note:
1. Each vertex is weighted by the length of the opposite side.
2. Check: the inradius of a 3-4-5 triangle is 1, so the incentre is 1 unit from each leg (x = 3, y = 2).

Answer: (c) 2 and 1

Explanation:
1. \( \overrightarrow{CA}=(2-a)\hat i+2\hat j \) and \( \overrightarrow{CB}=(1-a)\hat i-6\hat k \).
2. Right angle at C: \( \overrightarrow{CA}\cdot\overrightarrow{CB}=(2-a)(1-a)=0 \).
3. So a = 2 or a = 1.

Teacher's Note:
1. A right angle at C means CA · CB = 0.
2. Both roots give a valid (non-degenerate) triangle.