A-Level Scalar and vector products

Step 1: use the condition |a| = 2|b| to pin down p. |b| = sqrt(1² + (-2)² + 2²) = sqrt(1 + 4 + 4) = sqrt9 = 3 a = p(5i - 2j + 4k) with p > 0, so p comes out of the modulus as a positive factor: |a| = p sqrt(5² + (-2)² + 4²) = p sqrt(25 + 4 + 16) = p sqrt45 = 3 sqrt5 p Now impose |a| = 2|b|: 3 sqrt5 p = 2 x 3 = 6 p = 6/(3 sqrt5) = 2/sqrt5 = 2 sqrt5 / 5 (rationalising the denominator). Step 2: compute the cross product, keeping p outside as a scalar factor. a x b = p (5, -2, 4) x (1, -2, 2) Use the component rule (u x v)_1 = u2 v3 - u3 v2, and so on: i-component: (-2)(2) - (4)(-2) = -4 + 8 = 4 j-component: -[(5)(2) - (4)(1)] = -(10 - 4) = -6 k-component: (5)(-2) - (-2)(1) = -10 + 2 = -8 So a x b = p(4, -6, -8) = 2p(2, -3, -4) Substituting 2p = 2 x (2 sqrt5 / 5) = 4 sqrt5 / 5: a x b = (4 sqrt5 / 5)(2, -3, -4). Step 3: turn the cross product into an area. For two vectors drawn from the same point, |u x v| is the area of the parallelogram they span, so the triangle having them as two sides has area (1/2)|u x v|. First find the magnitude: |(2, -3, -4)| = sqrt(2² + (-3)² + (-4)²) = sqrt(4 + 9 + 16) = sqrt29 |a x b| = (4 sqrt5 / 5) sqrt29 = (4/5) sqrt(5 x 29) = 4 sqrt145 / 5 In the diagram the sides of triangle OAC are OA = a and OC = 2b, so OA x OC = 2(a x b) and area of triangle OAC = (1/2)|OA x OC| = (1/2) x 2|a x b| = |a x b| = 4 sqrt145 / 5.

Step 1: turn "b is perpendicular to 2a - b" into an equation. Two non-zero vectors are perpendicular exactly when their scalar product is zero, so b.(2a - b) = 0 Step 2: expand using the distributive property of the scalar product, and the fact that b.b = |b|². b.(2a) - b.b = 0 2(a.b) - |b|² = 0 2(a.b) = |b|² Step 3: replace a.b by its definition a.b = |a||b|cos alpha, where alpha is the angle between a and b. 2|a||b|cos alpha = |b|² Since b is non-zero, |b| ≠ 0, so divide both sides by |b|: 2|a|cos alpha = |b| Step 4: substitute the given relation |b| = sqrt(2)|a| and solve. 2|a|cos alpha = sqrt(2)|a| Since a is non-zero, |a| ≠ 0, so divide both sides by |a|: 2cos alpha = sqrt(2) cos alpha = sqrt(2)/2 = 1/sqrt(2) alpha is acute, so the only solution is alpha = pi/4 (that is, 45°)

Step 1 - see why OABC is a trapezium and which sides are the parallel ones. CB = OB - OC = b - c, and the question has already established b - c = lambda a. A vector that is a scalar multiple of another is parallel to it, so CB is parallel to OA. OA and CB are therefore the two parallel sides of the trapezium, of lengths |a| = 3 and BC. Step 2 - find the perpendicular height between those parallel sides. The height is the part of OC that is perpendicular to the direction of OA. Drop a perpendicular from C onto the line OA. In the right-angled triangle formed, the hypotenuse is OC of length |c| = 5 and the angle at O between a and c is 30 degrees. height = |c| sin 30 degrees = 5 x (1/2) = 5/2 units Step 3 - write down the area formula for a trapezium and substitute. area = (1/2) x (sum of the two parallel sides) x (perpendicular height) 10 = (1/2)(3 + BC)(5/2) Step 4 - solve for BC. 10 = (5/4)(3 + BC) Multiply both sides by 4: 40 = 5(3 + BC) Divide both sides by 5: 8 = 3 + BC BC = 8 - 3 = 5 BC = 5 units, as required.