Gce Model Questions Vol1 2024 Further Mathematics Question 41

Practice objective / multiple choice question 41 from the 2024 Gce Model Questions Vol1 Further Mathematics examination.

Gce Model Questions Vol1 2024 Further Mathematics Objective / Multiple Choice Medium Difficulty
Question 41 GCE_MODEL_QUESTIONS_VOL1 • 2024 • FURTHER_MATHEMATICS • objective

The vectors \(\mathbf{a} = (2,-1,3)\) and \(\mathbf{b} = (k,4,-2)\) are in three‑dimensional space. It is known that the magnitude of the projection of \(\mathbf{b}\) onto \(\mathbf{a}\) is \(5\). (i) Find all possible values of \(k\). (ii) For each value of \(k\), determine the magnitude of the cross‑product \(|\mathbf{a}\times\mathbf{b}|\).

Answer Options

A. k = 5 \pm \frac{5}{2}\sqrt{14}; \(|\mathbf{a}\times\mathbf{b}| = \sqrt{1600+350\sqrt{14}}\) for the plus sign and \(|\mathbf{a}\times\mathbf{b}| = \sqrt{1400-350\sqrt{14}}\) for the minus sign.
B. k = 5 \pm \frac{5}{3}\sqrt{14}; \(|\mathbf{a}\times\mathbf{b}| = \sqrt{1505+233\sqrt{14}}\) for the plus sign and \(|\mathbf{a}\times\mathbf{b}| = \sqrt{1505-233\sqrt{14}}\) for the minus sign.
Correct Answer C. k = 5 \pm \frac{5}{2}\sqrt{14}; \(|\mathbf{a}\times\mathbf{b}| = \sqrt{1505+350\sqrt{14}}\) for the plus sign and \(|\mathbf{a}\times\mathbf{b}| = \sqrt{1505-350\sqrt{14}}\) for the minus sign.
D. k = 5 \pm \frac{5}{2}\sqrt{7}; \(|\mathbf{a}\times\mathbf{b}| = \sqrt{1505+350\sqrt{7}}\) for the plus sign and \(|\mathbf{a}\times\mathbf{b}| = \sqrt{1505-350\sqrt{7}}\) for the minus sign.
Correct Answer
C
Correct Option:
k = 5 \pm \frac{5}{2}\sqrt{14}; \(|\mathbf{a}\times\mathbf{b}| = \sqrt{1505+350\sqrt{14}}\) for the plus sign and \(|\mathbf{a}\times\mathbf{b}| = \sqrt{1505-350\sqrt{14}}\) for the minus sign.

About This Question

This is Gce Model Questions Vol1 2024 Further Mathematics Question 41. It is one of the objective questions from the 2024 Gce Model Questions Vol1 Further Mathematics examination.

Difficulty level: Medium .

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