This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Step 1: Identify the two vectors. Let A = -2mi + 2j - 2k and B = 3i + 4mj + 3k. Step 2: Apply the condition for perpendicular vectors. Two vectors are perpendicular if their dot product is zero (A · B = 0). (-2m)(3) + (2)(4m) + (-2)(3) = 0 Step 3: Simplify the equation. -6m + 8m - 6 = 0 2m - 6 = 0 Step 4: Solve for m. 2m = 6 m = (6)/(2) m = 3 The value of m is 3. --- Step 1: Find the resultant vector R. The resultant vector is the sum of a_1 and a_2. R = a_1 + a_2 = (6i - 3j + 2k) + (3i + j - 2k) R = (6+3)i + (-3+1)j + (2-2)k R = 9i - 2j + 0k Step 2: Calculate the magnitude of the resultant vector |R|. |R| = sqrt(9^2 + (-2)^2 + 0^2) |R| = sqrt(81 + 4 + 0) |R| = sqrt(85) Step 3: Calculate the unit vector parallel to R. The unit vector u is R|R|. u = 9i - 2jsqrt(85) u = (9)/(sqrt(85))i - (2)/(sqrt(85))j The unit vector parallel to the resultant is (9)/(sqrt(85))i - (2)/(sqrt(85))j. --- Step 1: Identify the two vectors. Let A = 3i - 4j + 12k and B = i + 2j - k. Step 2: Calculate the cross product C = A × B. C = i & j & k \\ 3 & -4 & 12 \\ 1 & 2 & -1 C = i((-4)(-1) - (12)(2)) - j((3)(-1) - (12)(1)) + k((3)(2) - (-4)(1)) C = i(4 - 24) - j(-3 - 12) + k(6 + 4) C = -20i + 15j + 10k Step 3: Calculate the magnitude of C. |C| = sqrt((-20)^2 + 15^2 + 10^2) |C| = sqrt(400 + 225 + 100) |C| = sqrt(725) |C| = sqrt(25 × 29) = 5sqrt(29) Step 4: Calculate the unit vector perpendicular to A and B. The unit vector u is C|C|. u = -20i + 15j + 10k5sqrt(29) u = (-20)/(5sqrt(29))i + (15)/(5sqrt(29))j + (10)/(5sqrt(29))k u = -(4)/(sqrt(29))i + (3)/(sqrt(29))j + (2)/(sqrt(29))k The unit vector perpendicular to the given vectors is -(4)/(sqrt(29))i + (3)/(sqrt(29))j + (2)/(sqrt(29))k. --- i. Magnitude of a vector The magnitude of a vector is its length or size, representing the scalar value of the vector quantity. For a vector v = xi + yj + zk, its magnitude is given by |v| = sqrt(x^2 + y^2 + z^2). ii. Directional Cosines Directional cosines are the cosines of the angles that a vector makes with the positive x, y, and z axes. For a vector v = xi + yj + zk, the directional cosines are = (x)/(|v)|, = (y)/(|v)|, and = (z)/(|v)|. iii. Parallelism of two vectors Two vectors are parallel if they point in the same or opposite direction. Mathematically, two non-zero vectors A and B are parallel if one is a scalar multiple of the other, i.e., A = kB for some scalar k. Send me the next one 📸