Gce Model Questions Vol1 2025 Further Mathematics Question 34

Practice objective / multiple choice question 34 from the 2025 Gce Model Questions Vol1 Further Mathematics examination.

Gce Model Questions Vol1 2025 Further Mathematics Objective / Multiple Choice Hard Difficulty
Question 34 GCE_MODEL_QUESTIONS_VOL1 • 2025 • FURTHER_MATHEMATICS • objective

In three‑dimensional space let \(\mathbf{a} = (2,\,-1,\,k)\) and \(\mathbf{b} = (1,\,3,\,4)\). Find the value of \(k\) such that the angle between \(\mathbf{a}\) and \(\mathbf{b}\) is \(60^{\circ}\) and the magnitude of their cross product \(|\mathbf{a}\times\mathbf{b}|\) equals \(10\). Then give the unit vector in the direction of \(\mathbf{a}+\mathbf{b}\).

Answer Options

A. k = \dfrac{1+8/\sqrt{3}}{4}=\dfrac{\sqrt{3}+8}{4\sqrt{3}}. Unit vector \(\displaystyle \hat{u}=\frac{1}{\sqrt{\frac{1500+300\sqrt{3}}{48}}}\,(3,\;2,\;k+4)\).
Correct Answer B. k = \dfrac{1+10/\sqrt{3}}{4}=\dfrac{\sqrt{3}+10}{4\sqrt{3}}. Unit vector \(\displaystyle \hat{u}=\frac{1}{\sqrt{\frac{1591+340\sqrt{3}}{48}}}\,(3,\;2,\;k+4)\).
C. k = \dfrac{1+10/\sqrt{3}}{4}=\dfrac{\sqrt{3}+10}{4\sqrt{3}}. Unit vector \(\displaystyle \hat{u}=\frac{1}{\sqrt{\frac{1591+340\sqrt{3}}{48}}}\,(3,\;2,\;k+5)\).
D. k = \dfrac{1+12/\sqrt{3}}{4}=\dfrac{\sqrt{3}+12}{4\sqrt{3}}. Unit vector \(\displaystyle \hat{u}=\frac{1}{\sqrt{\frac{1700+360\sqrt{3}}{48}}}\,(3,\;2,\;k+4)\).
Correct Answer
B
Correct Option:
k = \dfrac{1+10/\sqrt{3}}{4}=\dfrac{\sqrt{3}+10}{4\sqrt{3}}. Unit vector \(\displaystyle \hat{u}=\frac{1}{\sqrt{\frac{1591+340\sqrt{3}}{48}}}\,(3,\;2,\;k+4)\).

About This Question

This is Gce Model Questions Vol1 2025 Further Mathematics Question 34. It is one of the objective questions from the 2025 Gce Model Questions Vol1 Further Mathematics examination.

Difficulty level: Hard .

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