Applications of Vector Algebra — Important Questions

9 units

Step 1: Find the resultant force: F = (2i+5j+6k) + (-i-2j-k) = i + 3j + 5k Step 2: Find displacement vector d = (6i+j-3k) - (4i-3j-2k) = 2i + 4j - k Step 3: Work done = F·d = (i+3j+5k)·(2i+4j-k) Step 4: = (1)(2) + (3)(4) + (5)(-1) = 2 + 12 - 5 = 9 Final: Total work done = 9 units. The key formula is W = F·d (dot product of force and displacement).

(x-1)/4 = (y-2)/5 = (z+3)/(-7)

Step 1: Identify the point on the line: (x1, y1, z1) = (1, 2, -3) Step 2: Identify direction ratios from the parallel vector 4i+5j-7k: b1=4, b2=5, b3=-7 Step 3: Apply the Cartesian equation formula: (x-x1)/b1 = (y-y1)/b2 = (z-z1)/b3 Step 4: Substitute: (x-1)/4 = (y-2)/5 = (z-(-3))/(-7) = (x-1)/4 = (y-2)/5 = (z+3)/(-7) Final: Note z+3 because z1 = -3, so z-z1 = z-(-3) = z+3. A common mistake is forgetting to change the sign.

π/2

Step 1: The acute angle θ between two lines with direction ratios (b1,b2,b3) and (d1,d2,d3) is given by cos θ = |b1d1+b2d2+b3d3| / (√(b1²+b2²+b3²) × √(d1²+d2²+d3²)) Step 2: Compute numerator: |(2)(4)+(1)(-4)+(-2)(2)| = |8-4-4| = |0| = 0 Step 3: Since the numerator is 0, cos θ = 0 Step 4: Therefore θ = cos⁻¹(0) = π/2 Final: The lines are perpendicular. When b1d1+b2d2+b3d3 = 0, the two lines are perpendicular to each other.

√365/3

Step 1: Identify a = 2i+3j+4k, b = -2i+j-2k, c = 3i-2k, d = 2i-j+2k. Note b = -1×(d), so lines are parallel! Step 2: For parallel lines: distance = |(c-a)×b| / |b| Step 3: c-a = (3-2)i+(0-3)j+(-2-4)k = i-3j-6k Step 4: (c-a)×b = |i j k / 1 -3 -6 / -2 1 -2| = i[(-3)(-2)-(-6)(1)] - j[(1)(-2)-(-6)(-2)] + k[(1)(1)-(-3)(-2)] = i[6+6] - j[-2-12] + k[1-6] = 12i+14j-5k Final: Distance = |12i+14j-5k| / |-2i+j-2k| = √(144+196+25) / √(4+1+4) = √365/3