Practice set
Vectors & Vector Operations II practice quiz with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
Which formula shows the distributive property of the dot product?
Correct answer: B. \(u\cdot(v+w) = u\cdot v + u\cdot w\)
Explanation: Distributive: \(u\cdot(v+w) = u\cdot v + u\cdot w\).
What is \(\cos\theta\) between vectors \((1,1,0)\) and \((1,2,3)\)?
Correct answer: C. \(\tfrac{3}{2\sqrt7}\)
Explanation: Compute dot = \(1+2+0=3\); magnitudes \(\sqrt2\) and \(\sqrt{14}\).
If \(u\) and \(v\) are unit and orthogonal, what is \(\|u+v\|\)?
Correct answer: B. \(\sqrt{2}\)
Explanation: Since perpendicular and unit: \(\sqrt{1^2 + 1^2} = \sqrt{2}\).
What is the distance from the point \((1,2,2)\) to the x-axis?
Correct answer: C. \(2\sqrt{2}\)
Explanation: Distance to the x-axis ignores the x-coordinate, so compute \(\sqrt{2^2 + 2^2} = 2\sqrt{2}\).
What is the projection of \((1,2,3)\) onto the xy-plane?
Correct answer: C. \((1,2,0)\)
Explanation: Drop z-component: \((1,2,0)\).
If \(u=(1,2,3)\) and \(v=(4,5,6)\), what is \((u×v)\cdot(1,1,1)\)?
Correct answer: D. \(0\)
Explanation: Compute \(u×v=(-3,6,-3)\), then dot with \((1,1,1)\) gives \(-3+6-3=0\).
What is the component of \((2,3,4)\) orthogonal to \((1,1,1)\)?
Correct answer: C. \((-1,0,1)\)
Explanation: Subtract projection \((3,3,3)\) from the original vector.
What is the component of \((1,2,3)\) perpendicular to \((3,0,0)\)?
Correct answer: B. \((0,2,3)\)
Explanation: Perpendicular = \((1,2,3) - (1,0,0) = (0,2,3)\).
What is the vector projection of \((2,2,1)\) onto \((1,1,0)\)?
Correct answer: A. \((2,2,0)\)
Explanation: Projection = \(\frac{u\cdot v}{u\cdot u}u = \frac{4}{2}(1,1,0) = (2,2,0)\).
What is the scalar triple product of \((1,2,3),(0,1,1),(2,0,1)\)?
Correct answer: B. \(-1\)
Explanation: Scalar triple product is the determinant of the matrix, which equals \(-1\).
Result
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Vectors & Vector Operations II Quiz, Explanations and Step-by-Step Review
Use the question set below to practise vectors and vector operations at the next level: the cross product in \(\mathbb{R}^3\) (including right-hand rule direction), area of a parallelogram and area of a triangle via \(\|u\times v\|\), the scalar triple product (also called the mixed product) for volume of a parallelepiped, coplanar vectors and the condition \((u\times v)\cdot w=0\), vector projection and scalar projection (components along a direction), distance from a point to a line/axis and distance from a point to a plane, and the Gram-Schmidt process to build an orthonormal basis. If you want a refresher with worked examples, click Start lesson.
How this vectors practice works
- 1. Take the practice set: answer the vectors and vector operations II questions below.
- 2. Open the lesson (optional): review cross product and triple product geometry, projection and scalar component, distance formulas, and Gram-Schmidt orthogonalization.
- 3. Retry: return to the question set and apply the correct vector formulas immediately.
What you will learn in the vectors & vector operations II lesson
Cross product & area in \(\mathbb{R}^3\)
- Cross product computation: \(u\times v\) component formula and determinant form
- Perpendicular vectors and the right-hand rule direction
- Area: \(\|u\times v\|\) (parallelogram) and \(\dfrac12\|u\times v\|\) (triangle)
Scalar triple product, determinants & volume
- Scalar triple product: \((u\times v)\cdot w=\det[u\;v\;w]\)
- Volume of a parallelepiped: \(\left|(u\times v)\cdot w\right|\)
- Coplanarity test: \((u\times v)\cdot w=0\) (volume \(=0\))
Projection, scalar component & distances
- Vector projection: \(\mathrm{proj}_b a=\dfrac{a\cdot b}{b\cdot b}\,b\) and scalar projection: \(\mathrm{comp}_b a=\dfrac{a\cdot b}{\|b\|}\)
- Distance to a line/axis: \(\|a-\mathrm{proj}_d a\|\) (or \(\dfrac{\|a\times d\|}{\|d\|}\))
- Distance to a plane using a normal vector: \(\dfrac{|n\cdot a-d|}{\|n\|}\)
Gram-Schmidt & orthonormal bases
- Gram-Schmidt process: build an orthogonal then orthonormal set
- Orthogonal component: subtract projections step-by-step
- Why it matters: clean coordinates, stable geometry, and foundations for QR decomposition

