Solution: A.B = ABcosθ = ABcos 80° = 0.174AB. When the angle is increased by 20° the new angle is 100°. So, A.B = ABcos100° = –  0.174AB.

Solution
MINIMUM NUMBER OF UNEQUAL VECTORS TO GIVE NULL VECTOR: On the minimum, three unequal vectors will give a zero resultant (or null vector).
The resultant is said to be zero or null if the vectors being added forms a close shape when put head to tail. So three vectors of unequal magnitudes when put head to tail and they form a triangle (triangle is a close shape), the resultant is a null vector.
In the diagram three vectors A ,B and C are added by head-to-tail method. The resultant vector is zero.

Note that two vectors can also give a zero resultant if they are equal and anti-parallel. However, the condition of the problem demands the vectors to be unequal. Therefore, we can’t take the case of two vectors.

Solution: See solution here.

Solution: The magnitude of vector A in space is

Here AX = 2, AY = 1, AZ = 2, therefore,

Solution:

Solution: See the diagram below.

Clockwise torque = 5.88 × 0.3 = 1.674 N m … (A)
Counterclockwise torque = 3.92 × 0.3 = 1.176 … (B)
Difference in torques = 1.764 – 1.176 = 0.588 N m
So we need 0.588 N m torque in the counterclockwise direction from a mass of 0.3 kg = 2.94 N. Therefore, if x is the distance from the point of balancing, then 2.94x = 0.588 ⇒ x = 0.588/2.94 = 0.2 m. Hence the third mass should be hung 0.2 m = 20 cm from the point of balancing. As the point of balancing is at 50 cm, therefore, the mass should be hung on 30 cm mark.

Explanation: Since the y-component is negative and x-component is positive, the vector points in the IV quadrant. (Draw the vector according to the scale).

Explanation: Let A × B = C. Now C is perpendicular to both A and B. Now as A.C = AC cosθ = AC cos90° and cos90° = 0. Therefore, option A is correct.

Explanation: If A and B are two vectors to be added, then depending on the angle between them, the magnitude of the resultant is either equal to or less than A + B and greater than or equal to A – B. Now the sum of magnitudes of the given vectors is 20 + 50 = 70 and difference is 50 -20 = 30. Therefore, the range of the sum is from 30 to 70 N. Only 20 N is out of this range. Therefore, A is the correct option.

Explanation: A.B = ABcosθ, and since A and B are nonzero, therefore, cosθ = 0 ⇒ θ = 90°. Now A X B = ABsinθ = ABsin90° = AB. (∵ sin90° = 1).

Solution: Let x₁ and x₂ are the magnitudes of two vectors such that x₁ is less than x₂. Now if the resultant force is perpendicular to x₁, then the three vectors when combined make a right angled triangle as shown in the figure.


We know that for a right angled triangle,
(Hypotenuse)² = (base)² + (perpendicular)².

Given in the problem is

Put this value of x₂ in equation (1)

Put x₁ = 6 in equation (2), we have x₂ = 16 – 6 = 10 N

SOLUTION: Clockwise torque produced by 80 g (= 0.08 kg) mass to the right = 20 × 9.8 × 0.08 = 15.68 N m.
Let m be the mass of the rod, then the anti-clockwise torque produced by the weight mg to the left = 9.8m × 16 = 156.8m N m. Since the rod is in equilibrium, both torques are equal.
156.8m = 15.68 OR m = 15.68/156.8 = 0.1 kg = 100 g.

Solution: We are concerned with the magnitudes of scalar and vector products of the two vectors. Since the scalar and vector multiplications differ only in the difference of the sine and cosine values of the angle between them, therefore, if the angle between the vectors is such that both its sine and cosine values are equal, the magnitudes of the scalar and vector product of such vectors will also be equal. Now, 45⁰ is the angle whose sine = cosine = 1/√2, therefore, if two vectors are making 45⁰ angle between them, the magnitudes of the scalar and vector products of such vectors will be equal. Here in this case,

Solution Hint: The force 10 N is producing a clockwise torque. Take it as negative. The 5 N force is producing anti-clockwise torque. Take it positive. Apply the formula and find both torques. Then add them. Remember the anti-clockwise torque is positive. Θ in both cases is 90^o.