Relative to a fixed origin O - the point A has position vector 4i - 3j + 5k - the point B has position vector 4j + 6k - the point C has position vector -16i + pj + 10k where p is a constant. Given that A, B and C lie on a straight line, a) find the value of p. The line segment OB is extended to a point D so that CD is parallel to OA. b) Find |OB|, writing your answer as a fully simplified surd.

A-Level Edexcel Maths Pure Modelling with Functions: Relative to a fixed origin O -

Relative to a fixed origin O - the point A has position vector 4i - 3j + 5k - the point B has position vector 4j + 6k - the point C has position vector -16i + pj + 10k where p is a constant - Edexcel - A-Level Maths Pure - Question 14 - 2022 - Paper 2

Question 14

Relative-to-a-fixed-origin-O----the-point-A-has-position-vector-4i---3j-+-5k---the-point-B-has-position-vector-4j-+-6k---the-point-C-has-position-vector--16i-+-pj-+-10k--where-p-is-a-constant-Edexcel-A-Level Maths Pure-Question 14-2022-Paper 2.png

Relative to a fixed origin O - the point A has position vector 4i - 3j + 5k - the point B has position vector 4j + 6k - the point C has position vector -16i + pj + ... show full transcript

Worked Solution & Example Answer:Relative to a fixed origin O - the point A has position vector 4i - 3j + 5k - the point B has position vector 4j + 6k - the point C has position vector -16i + pj + 10k where p is a constant - Edexcel - A-Level Maths Pure - Question 14 - 2022 - Paper 2

Step 1

a) find the value of p

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Answer

To determine the value of p such that points A, B, and C are collinear, we can use the position vectors:

Position vector of A: $ extbf{a} = 4i - 3j + 5k $
Position vector of B: $ extbf{b} = 0i + 4j + 6k $
Position vector of C: $ extbf{c} = -16i + pj + 10k $

Since A, B and C are collinear, vectors AB and AC must be parallel:

Vector AB: \( extbf{b} - extbf{a} = (0i + 4j + 6k) - (4i - 3j + 5k) = -4i + 7j + k\
Vector AC: \( extbf{c} - extbf{a} = (-16i + pj + 10k) - (4i - 3j + 5k) = -20i + (p + 3)j + 5k\

Since these vectors must be scalar multiples of each other, we can set up the equations based on their corresponding components:

For the i-component: \( -4 = -20k ext{ for some scalar } k\
For the j-component: $ 7 = (p + 3)k $
For the k-component: \( 1 = 5k \

From the k-component:

Solving for k: \( ightarrow k = \frac{1}{5}\

Substituting k back into the other equations:

From the i-component: \( -4 = -20 \cdot \frac{1}{5} = -4 \ (correct)
From the j-component: $ 7 = (p + 3) \cdot \frac{1}{5} $

Solving for p:

Multiplying through by 5: \( 5 imes 7 = p + 3 \
\( 35 = p + 3\
\( p = 35 - 3 = 32\

Thus, the value of p is 32.

Step 2

b) Find |OB|, writing your answer as a fully simplified surd.

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Answer

To find the magnitude of the vector OB:

Position vector of O: $ extbf{o} = 0i + 0j + 0k $
Position vector of B: $ extbf{b} = 0i + 4j + 6k $

Using the formula for the magnitude of a vector: $ | extbf{OB}| = \sqrt{(0-0)^2 + (4-0)^2 + (6-0)^2} $ = \ $|OB| = \\sqrt{0 + 16 + 36} = \\sqrt{52} = 2\sqrt{13} \\$ (fully simplified)

Thus, |OB| = ( 2\sqrt{13} ).

Relative to a fixed origin O - the point A has position vector 4i - 3j + 5k - the point B has position vector 4j + 6k - the point C has position vector -16i + pj + 10k where p is a constant - Edexcel - A-Level Maths Pure - Question 14 - 2022 - Paper 2

Worked Solution & Example Answer:Relative to a fixed origin O - the point A has position vector 4i - 3j + 5k - the point B has position vector 4j + 6k - the point C has position vector -16i + pj + 10k where p is a constant - Edexcel - A-Level Maths Pure - Question 14 - 2022 - Paper 2

a) find the value of p

b) Find |OB|, writing your answer as a fully simplified surd.

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