ベクトル計算機

Dot and cross products of two vectors in 2D or 3D, their magnitudes, the angle between them and the projection of one onto the other, with a drawing.

更新日 検証済みの例:5

試す
Dot product a · b
Dot product a · b: 32
小数点以下の最大桁数:10;最も近い値へ、等距離ならゼロから遠い値へ
Cross product a × b
(−3, 6, −3)
|a × b| (parallelogram area)
7.3484692283
|a|
3.7416573868
|b|
8.7749643874
Angle between a and b
12.93315449°
Angle in radians
0.2257261286rad
Scalar projection of a onto b
3.6467384467
Vector projection of a onto b
(1.662338, 2.077922, 2.493506)

a · b = 32, so the vectors are 12.9332° apart; the parallelogram they span has area 7.34847.

Vectors a, b and a × b (oblique view)

xyzaba × b
計算方法 S
  1. Dot product

    a⋅b=1⋅4+2⋅5+3⋅6=32\mathbf a\cdot\mathbf b = 1\cdot4 + 2\cdot5 + 3\cdot6 = 32
  2. Cross product

    a×b=(aybz−azbyazbx−axbzaxby−aybx)=(−36−3)\mathbf a\times\mathbf b = \begin{pmatrix} a_y b_z - a_z b_y \\ a_z b_x - a_x b_z \\ a_x b_y - a_y b_x \end{pmatrix} = \begin{pmatrix}-3 \\ 6 \\ -3\end{pmatrix}
  3. Magnitudes

    ∣a∣=12+22+32=3.74165738677,∣b∣=42+52+62=8.77496438739|\mathbf a| = \sqrt{1^2 + 2^2 + 3^2} = 3.74165738677,\qquad |\mathbf b| = \sqrt{4^2 + 5^2 + 6^2} = 8.77496438739
  4. Angle between them

    θ=atan2⁡(∣a×b∣, a⋅b)=atan2⁡(7.348469228, 32)=12.9331544919∘\theta = \operatorname{atan2}\left(|\mathbf a\times\mathbf b|,\ \mathbf a\cdot\mathbf b\right) = \operatorname{atan2}(7.348469228,\ 32) = 12.9331544919^\circ

    Equivalent to arccos(a·b / |a||b|), but stays accurate when the vectors are nearly parallel.

  5. Projection of a onto b

    comp⁡ba=a⋅b∣b∣=3.64673844671,proj⁡ba=3277 b=(1.6623376622.0779220782.493506494)\operatorname{comp}_{\mathbf b}\mathbf a = \frac{\mathbf a\cdot\mathbf b}{|\mathbf b|} = 3.64673844671,\qquad \operatorname{proj}_{\mathbf b}\mathbf a = \frac{32}{77}\,\mathbf b = \begin{pmatrix}1.662337662 \\ 2.077922078 \\ 2.493506494\end{pmatrix}

ベクトル計算機について

The dot product multiplies matching components and adds them, a · b = a₁b₁ + a₂b₂ + a₃b₃. The cross product of two 3D vectors is a vector perpendicular to both, and its length equals the area of the parallelogram they span. The angle between the vectors is computed as θ = atan2(|a × b|, a · b), which equals arccos(a · b / |a||b|) but stays accurate for nearly parallel vectors, and the projection of a onto b is (a · b / |b|²) b.

Physics (work and torque), 3D graphics and linear algebra are the usual uses. The default vectors a = (1, 2, 3) and b = (4, 5, 6) give a · b = 32, a × b = (−3, 6, −3) and an angle of about 12.93°.

A dot product of 0 means the vectors are perpendicular; a zero cross product means they are parallel or one is zero. In 2D mode both vectors lie in the xy-plane, so their cross product points along z: (3, 4) × (4, −3) = (0, 0, −25).

計算例

a = (1, 2, 3), b = (4, 5, 6)

Dimensions
3D
a — x
1
a — y
2
a — z
3
b — x
4
b — y
5
b — z
6
Dot product a · b
32
Cross product a × b
(−3, 6, −3)
|a|
3.7416573868
Angle between a and b
12.93315449 °
|a × b| (parallelogram area)
7.3484692283

照合元:Hand calculation; Python decimal √14, √54 and atan2(√54, 32) in degrees (hp.py)

Perpendicular 2D vectors (3, 4) and (4, −3)

Dimensions
2D
a — x
3
a — y
4
b — x
4
b — y
-3
Dot product a · b
0
Angle between a and b
90 °
Cross product a × b
(0, 0, −25)

照合元:3·4 + 4·(−3) = 0; 3·(−3) − 4·4 = −25

Parallel vectors (edge case)

Dimensions
3D
a — x
1
a — y
2
a — z
3
b — x
2
b — y
4
b — z
6
Cross product a × b
(0, 0, 0)
Angle between a and b
0 °
Dot product a · b
28

照合元:b = 2a, so a × b = 0 and θ = 0

Projection of (2, 3) onto (4, 0)

Dimensions
2D
a — x
2
a — y
3
b — x
4
b — y
0
Scalar projection of a onto b
2
Vector projection of a onto b
(2, 0)
Angle between a and b
56.30993247 °

照合元:a·b/|b| = 8/4; angle atan2(3, 2) = 56.3099324740202…° (Python math.degrees)

よくある質問

How do you calculate the dot product of two vectors?

Multiply corresponding components and add the results: a · b = a₁b₁ + a₂b₂ + a₃b₃. For (1, 2, 3) · (4, 5, 6) that is 4 + 10 + 18 = 32. The dot product also equals |a||b| cos θ, so it is positive when the angle is under 90°, 0 at exactly 90° and negative beyond it: (1, 0) · (−2, 0) = −2.

How do you calculate the cross product?

For a = (a₁, a₂, a₃) and b = (b₁, b₂, b₃), a × b = (a₂b₃ − a₃b₂, a₃b₁ − a₁b₃, a₁b₂ − a₂b₁). For (1, 2, 3) × (4, 5, 6) that gives (12 − 15, 12 − 6, 5 − 8) = (−3, 6, −3). The result is perpendicular to both vectors, follows the right-hand rule, and changes sign if the order is swapped: b × a = (3, −6, 3).

How do you find the angle between two vectors?

Use cos θ = (a · b)/(|a||b|). For (1, 2, 3) and (4, 5, 6), cos θ = 32/(√14 × √77) ≈ 0.974632, so θ ≈ 12.93°. Near 0° or 180° the arccos form loses accuracy, so the calculator uses the equivalent θ = atan2(|a × b|, a · b), the form William Kahan recommends in his notes on floating-point roundoff.

What does it mean if the dot product is zero?

The two vectors are perpendicular (orthogonal), provided neither is the zero vector. (3, 4) · (4, −3) = 12 − 12 = 0, so those vectors meet at exactly 90°. In physics the same test shows that a force at right angles to the motion does no work, since work is the dot product of force and displacement.

What is the difference between scalar and vector projection?

The scalar projection of a onto b is the signed length of a along b, a · b / |b|; the vector projection is that length times the unit vector of b, (a · b / |b|²) b. Projecting (2, 3) onto (4, 0) gives a scalar projection of 8/4 = 2 and a vector projection of (2, 0). A negative scalar projection means a points partly against b.

「ベクトル計算機」の精度はどのくらいですか?

精度は入力値と計算方法の前提に依存します。十進演算には有効数字50桁を使いますが、推定、数値計算手法、元データの精度はそれより低い場合があります。表示の丸め処理でこれらの制約がなくなるわけではありません。 独立した出典の解答と照合した計算例:5。 例えば、「a = (1, 2, 3), b = (4, 5, 6)」はHand calculation; Python decimal √14, √54 and atan2(√54, 32) in degrees (hp.py)と照合しています。

この計算方法の出典は何ですか?

Wolfram MathWorld — Dot Product; Wolfram MathWorld — Cross Product; W. Kahan, How futile are mindless assessments of roundoff in floating-point computation? §12 — angles via atan2 rather than arccos.

この計算機について

a⋅b=∑aibiθ=atan2⁡(∣a×b∣, a⋅b)proj⁡ba=a⋅b∣b∣2 b\begin{gathered} \mathbf a\cdot\mathbf b = \sum a_i b_i \\[6pt] \theta = \operatorname{atan2}(|\mathbf a\times\mathbf b|,\ \mathbf a\cdot\mathbf b) \\[6pt] \operatorname{proj}_{\mathbf b}\mathbf a = \frac{\mathbf a\cdot\mathbf b}{|\mathbf b|^2}\,\mathbf b \end{gathered}

出典

  1. Wolfram MathWorld — Dot Product
  2. Wolfram MathWorld — Cross Product
  3. W. Kahan, How futile are mindless assessments of roundoff in floating-point computation? §12 — angles via atan2 rather than arccos

出典と照合済み

この計算機には、独立した出典の解答を使った計算例が 5 件あります。テストに組み込まれており、ここでも実行できます。

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