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VICSpecialist MathematicsQuick questions

Unit 3: Vectors

Quick questions on Vectors operations and applications: VCE Specialist Mathematics Unit 3

8short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.

What are magnitudes?
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∣a∣=22+12+(−2)2=4+1+4=9=3|\mathbf{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3. For b=3i+0j−4k\mathbf{b} = 3\mathbf{i} + 0\mathbf{j} - 4\mathbf{k}, ∣b∣=32+02+(−4)2=9+16=25=5|\mathbf{b}| = \sqrt{3^2 + 0^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.
What is dot product?
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a⋅b=(2)(3)+(1)(0)+(−2)(−4)=6+0+8=14\mathbf{a} \cdot \mathbf{b} = (2)(3) + (1)(0) + (-2)(-4) = 6 + 0 + 8 = 14.
What is angle?
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cos⁡θ=143×5=1415\cos\theta = \dfrac{14}{3 \times 5} = \dfrac{14}{15}, so θ=cos⁡−1 ⁣(1415)≈21.0∘\theta = \cos^{-1}\!\left(\frac{14}{15}\right) \approx 21.0^\circ.
What is vector resolute?
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Using proj⁡ba=a⋅b∣b∣2b=1425(3i−4k)\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\mathbf{b} = \dfrac{14}{25}(3\mathbf{i} - 4\mathbf{k}), we get
What is sign of the scalar resolute?
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If the angle is obtuse the scalar resolute is negative. Do not take the absolute value unless a length is explicitly requested.
What is q1?
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Find the angle between a=i+j\mathbf{a} = \mathbf{i} + \mathbf{j} and b=i\mathbf{b} = \mathbf{i}. [2 marks]
What is q2?
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Find the scalar resolute of a=2i−j+2k\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k} in the direction of b=i+2j+2k\mathbf{b} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}. [2 marks]
What is q3?
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Determine the value of tt for which a=ti+3j\mathbf{a} = t\mathbf{i} + 3\mathbf{j} is perpendicular to b=2i−j\mathbf{b} = 2\mathbf{i} - \mathbf{j}. [2 marks]

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