A-Level Edexcel Maths Pure Statistical Measures: Figure 1 shows a sketch of trian

Figure 1 shows a sketch of triangle ABC - Edexcel - A-Level Maths Pure - Question 7 - 2021 - Paper 1

Question 7

Figure 1 shows a sketch of triangle ABC. Given that • AB = -3i - 4j - 5k • BC = i + j + 4k (a) find AC (b) show that cosABC = \frac{9}{10}

Worked Solution & Example Answer:Figure 1 shows a sketch of triangle ABC - Edexcel - A-Level Maths Pure - Question 7 - 2021 - Paper 1

Step 1

96%

114 rated

Answer

To find the vector \vec{AC}, we can use the relationship:

\vec{AC} = \vec{AB} + \vec{BC}

Given:

Now we compute \vec{AC}:

\vec{AC} = (-3i - 4j - 5k) + (i + j + 4k)

Combining the like terms:

Thus, we have:

\vec{AC} = -2i - 3j - 1k

Step 2

99%

104 rated

Answer

To find ( \cos \angle ABC ), we can use the formula:

\cos \theta = \frac{\vec{AB} \cdot \vec{BC}}{||\vec{AB}|| \cdot ||\vec{BC}||}

First, we need to calculate the dot product ( \vec{AB} \cdot \vec{BC} ):

\vec{AB} \cdot \vec{BC} = (-3)(1) + (-4)(1) + (-5)(4) = -3 - 4 - 20 = -27

Next, we find the magnitudes of both vectors:

||\vec{AB}|| = \sqrt{(-3)^2 + (-4)^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}

||\vec{BC}|| = \sqrt{(1)^2 + (1)^2 + (4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}

Now substituting back into the formula:

\cos \angle ABC = \frac{-27}{(5\sqrt{2})(3\sqrt{2})} = \frac{-27}{30} = -\frac{9}{10}

Since the angle is acute, we take the positive value:

\cos ABC = \frac{9}{10}