Vectors (Further Pure 1)
Introduced| English | Meaning |
|---|---|
| plane/pleɪn/ | A flat two-dimensional surface extending indefinitely in every direction within it. |
| vector product/ˈvektə ˈprɒdʌkt/ | The cross product of two three-dimensional vectors, perpendicular to both with magnitude equal to their magnitudes times the sine of the angle between them. |
| scalar product/ˈskeɪlə ˈprɒdʌkt/ | The scalar found by adding the products of corresponding vector components; for non-zero vectors it is their magnitude product times the cosine of the included angle. |
| perpendicular/ˌpɜːpənˈdɪkjʊlə/ | Meeting at a right angle of 90 degrees. |
| anti-commutative/ˌæntikəˈmjuːtətɪv/ | Having a product whose sign reverses when its two inputs are exchanged. |
| parallelogram/ˌpærəˈleləɡræm/ | A quadrilateral with both pairs of opposite sides parallel. |
| unit normal/ˈjuːnɪt ˈnɔːml/ | A normal vector divided by its magnitude, so that it has length one. |
| normal vector/ˈnɔːml ˈvektə/ | A non-zero vector perpendicular to all directions in a plane. |
| foot of a perpendicular/fʊt əvə ˌpɜːpənˈdɪkjʊlə/ | The point where the perpendicular from a given point meets a line or plane. |
| skew/skjuː/ | Describing two lines in three-dimensional space that are not parallel and do not intersect. |
| common perpendicular/ˈkɒmən ˌpɜːpənˈdɪkjʊlə/ | A line that intersects both given lines at right angles; for skew lines it contains their shortest connecting segment. |
Find the shortest safe route in space
- A drone must approach a flat roof by the shortest route. A cable must join two rails that do not meet. These are different geometry problems: first identify the points, lines and plane 平面 involved.
- A direction alone is not a complete line. A distance alone does not locate its endpoints. Vector products 向量积 and scalar products 数量积 let you construct the geometry and then check it.
Choose the product that answers the question
- The scalar product is $\mathbf u\cdot\mathbf v=u_1v_1+u_2v_2+u_3v_3$. For non-zero vectors it equals $|\mathbf u||\mathbf v|\cos\theta$, so a zero scalar product identifies perpendicular 垂直 directions. The zero vector has no direction or defined angle.
- The vector product is anti-commutative 反交换: swapping the order reverses its sign. Its magnitude is $|\mathbf u||\mathbf v|\sin\theta$, the area of the parallelogram 平行四边形 spanned by the vectors. A triangle has half this area. For non-parallel inputs, $\hat{\mathbf n}$ is the unit normal 单位法向量 in the ordered product direction.
For $\mathbf u=(1,1,0)$ and $\mathbf v=(0,1,1)$,
The cross product has zero scalar product with each input and magnitude $\sqrt3$. Parallel vectors have zero cross product, so they cannot supply a normal direction to a unique plane. For comparison, $(1,1,0)$ and $(1,-1,0)$ have zero scalar product and are perpendicular.
Check perpendicular directions in a plane
a · b = |a||b|cos θ
For non-zero vectors, a zero scalar product identifies perpendicular directions. Compare (1,1) with (1,−1), then vary a component.
For two non-parallel non-zero vectors a and b, their vector product a×b is:
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a × b is perpendicular to both vectors; its length is |a||b|sin θ.
The magnitude |a × b| equals the:
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|a × b| = |a||b|sin θ, which is the parallelogram's area.
The vector product is commutative: a × b = b × a.
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The vector product is anti-commutative: a × b = −(b × a). Swapping the order reverses the direction.
If |a| = 3, |b| = 4, and the angle between them is 30°, |a × b| = |a||b|sin θ. Find it.
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|a × b| = 3 × 4 × sin 30° = 12 × 0.5 = 6.
Two vectors have magnitudes 5 and 6 with an angle of 90 degrees between them. What is the magnitude of their vector product, and what happens if the order is swapped?
The magnitude of a cross product is the product of the magnitudes times the sine of the included angle. The product is anti-commutative.
For u=(2,−1,1) and v=(1,2,0), calculate u×v and u·v. Find the ordered unit normal, the parallelogram area and the triangle area. State what changes when the order of the cross product is reversed.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
For u=(1,−2,3) and v=(2,−4,6), find both products and explain whether they provide a normal for a unique plane. What changes if u is instead the zero vector? Does a zero scalar product then give a defined angle of 90°?
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
Read the three forms of a plane
- A normal vector 法向量 is a non-zero vector perpendicular to every direction in the plane. The Cartesian form $ax+by+cz=d$ is the scalar-product form $\mathbf r\cdot\mathbf n=d$, with $\mathbf n=(a,b,c)$. A unit normal is $\mathbf n/|\mathbf n|$. The number $d$ is not generally the distance from the origin unless the normal is a unit vector.
- In $\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c$, $\mathbf a$ locates a point and the non-parallel vectors $\mathbf b,\mathbf c$ give directions in the plane. The parameters vary independently. Parallel direction vectors describe only a line, not a plane.
A plane with normal $(2,-1,3)$ through $(1,0,2)$ has
Multiplying every coefficient, including the right side, by the same non-zero constant gives the same plane.
A plane has equation 2x − y + 3z = 8. The normal vector has components (2, −1, ?). Find the missing component.
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The coefficients of x, y, z in ax + by + cz = d give the normal vector (a, b, c) = (2, −1, 3).
Convert a parametric plane to Cartesian form
- Take the cross product of its two direction vectors to find a normal. Then use a known point to find the constant: $\mathbf n\cdot\mathbf r=\mathbf n\cdot\mathbf a$. For three given points, first form two displacement vectors from one of them.
- To return to a parametric equation, choose two coordinates as free parameters, solve for the third and separate a fixed point from the two direction terms. Check both directions have zero scalar product with the normal.
For
the normal is $(1,-1,1)$, giving $x-y+z=3$. Conversely, set $x=1+\lambda$ and $z=2+\mu$. The Cartesian equation gives $y=\lambda+\mu$, recovering the original parametric form.
Convert r=(2,−1,1)+λ(1,2,0)+μ(0,1,2) to Cartesian and scalar-product forms. Recover a parametric form from your Cartesian answer and check its point and direction vectors.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
Find the Cartesian equation of the plane through A=(1,1,0), B=(3,1,1), C=(1,2,2), and the area of triangle ABC. Explain why crossing two position vectors would not generally be the right construction.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
A=(0,0,0), B=(1,1,0), C=(2,k,0). Find the values of k for which these points determine a unique plane, give that plane, and find the triangle area. What happens at the excluded value?
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
Classify a line and a plane before solving
- Substitute $\mathbf r=\mathbf a+t\mathbf u$ into $\mathbf n\cdot\mathbf r=d$. If $\mathbf n\cdot\mathbf u\ne0$, there is one intersection, found from the parameter below. Substitute back to give the actual point.
- If $\mathbf n\cdot\mathbf u=0$, test the base point. When $\mathbf n\cdot\mathbf a=d$, the whole line lies in the plane; otherwise it is parallel and separate. Dividing by zero would hide this distinction.
For the plane $x-y+z=3$, the line $\mathbf r=t(1,0,1)$ meets it when $2t=3$, at $(3/2,0,3/2)$. The line $(1,0,2)+t(1,1,0)$ lies in the plane. The line $t(1,1,0)$ is parallel and separate, because its points satisfy $x-y+z=0$ instead.
Classify each line relative to 2x+y −z=4: L₁:r=t(1,0,1), L₂:r=(2,0,0)+t(0,1,1), L₃:r=t(0,1,1). Find any unique intersection point and show why the other cases differ.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
Find the foot of a perpendicular 垂足
- The foot $H$ from a point $P$ to a plane lies on the normal line through $P$. Write $\mathbf h=\mathbf p-t\mathbf n$, substitute into the plane equation, then solve for $t$. The sign determines which way to travel along the normal.
- Verify both conditions: $H$ lies in the plane and $\overrightarrow{PH}$ is parallel to its normal. Merely finding some point in the plane does not make it the nearest point.
From $P=(3,0,3)$ to $x-y+z=3$, use $\mathbf n=(1,-1,1)$:
Thus $t=1$ and $H=(2,1,2)$. It satisfies $2-1+2=3$, and $\overrightarrow{PH}=(-1,1,-1)$ is parallel to the normal.
Measure a point-to-plane distance
- The shortest distance is the component of the displacement in the normal direction. For a non-zero normal, use the absolute numerator to obtain a non-negative distance; do not confuse the plane constant with the distance symbol.
- Multiplying a plane equation by a non-zero constant changes numerator and denominator by the same magnitude, leaving the distance unchanged. A zero numerator means the point already lies in the plane.
For $P=(3,0,3)$ and $x-y+z=3$, $D=|6-3|/\sqrt3=\sqrt3$. This agrees with the length of $\overrightarrow{PH}$ above. The plane's distance from the origin is also $|3|/\sqrt3=\sqrt3$, although these are different point-to-plane problems.
Find the foot H of the perpendicular from P=(4,−1,3) to 2x −y+2z=6. Find the shortest distance and verify both geometric conditions on H.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
The equations 2x −y+2z=6 and −4x+2y −4z=−12 describe one plane. Find its distance from the origin using both equations. Is either right-hand constant itself the distance? Give a unit-normal form.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
Use normals for plane angles
- The smaller angle between two planes is the smaller angle between their normal directions, with $0\le\alpha\le\pi/2$. The absolute scalar product removes the choice of normal direction: $\mathbf n$ and $-\mathbf n$ describe the same plane.
- The angle $\phi$ between a line and a plane is complementary to the angle between its direction and the plane normal. Use sine for the line–plane angle, with $0\le\phi\le\pi/2$, and check both direction vectors are non-zero.
The planes $x-y+z=3$ and $x+y+z=5$ have normals with scalar product $1$ and lengths $\sqrt3$, so $\alpha=\arccos(1/3)\approx70.5^\circ$. A line with direction $(1,0,1)$ makes angle $\phi=\arcsin(2/\sqrt6)\approx54.7^\circ$ with the first plane. Using cosine would give its angle with the normal instead.
Find the acute angle between the planes with normals(1,2,2) and(2,−1,2). Also find the angle between a line of direction(2,1,2) and the first plane. Explain why the inverse trigonometric functions differ.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
Find the intersection of two planes
- For non-parallel normals, the intersection direction is $\mathbf n_1\times\mathbf n_2$. Find a common point by solving both plane equations, then give a full line equation: point plus parameter times direction.
- If the normals are parallel, compare the constants using the same scale factor: the planes are either identical or parallel and separate. A zero cross product cannot provide an intersection direction.
Subtract $x-y+z=3$ from $x+y+z=5$ to get $y=1$, then $x+z=4$. A common point is $(4,1,0)$. The cross product of the normals is $(-2,0,2)$, so an equivalent intersection line is
Every point of this line satisfies both plane equations; there is no need to use the unsimplified direction vector.
Find a vector equation of the intersection of x+2y −z=3 and 2x −y+z=6. Show that the point and direction satisfy both planes, and explain why a cross product alone is incomplete.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
Compare Π:x+2y −z=3 with Σ:2x+4y −2z=6 and T:2x+4y −2z=7. Which pairs have an intersection line? Find the distance from Π to T.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
Decide whether two lines are skew 异面
- Skew lines are neither parallel nor intersecting. For $\mathbf r=\mathbf a+s\mathbf u$ and $\mathbf r=\mathbf b+t\mathbf v$, compare their directions and, if non-parallel, solve all three coordinate equations for an intersection. Two different line parameters must be used.
- When $\mathbf u\times\mathbf v\ne0$, the lines are coplanar exactly when $(\mathbf b-\mathbf a)\cdot(\mathbf u\times\mathbf v)=0$. A non-zero value proves they are skew. Do not apply the skew-distance denominator to parallel lines.
This schematic uses the x-axis and a y-parallel line at height 2. Their nearest points have the same x and y coordinates, so their connector has length 2.
Classify L₁:r=(1,0,2)+s(1,1,0) and L₂:r=(2,1,2)+t(0,1,1). Find their intersection or show that none exists. Check the result with the scalar triple product.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
Calculate the shortest skew-line distance
- For non-parallel lines, $\mathbf u\times\mathbf v$ is perpendicular to both. Project the displacement between their base points onto this direction. Any non-zero multiple of the cross product gives the same distance.
- The formula below gives a distance, not the location of the shortest segment. For parallel lines use $|(\mathbf b-\mathbf a)\times\mathbf u|/|\mathbf u|$ instead; for intersecting non-parallel lines the numerator is zero.
Let
Their cross product is $(1,-1,1)$ and their base-point displacement is $(1,2,2)$. Its scalar product with the cross product is $1$, so they are skew and $D=1/\sqrt3$.
Scalar and vector products can find the shortest distance between two skew lines.
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These products let you compute distances, angles and intersections in 3D, including skew-line distances.
L₁ is the x-axis and L₂:r=(0,1,k)+t(0,1,1). Find every real k giving shortest distance√2 between them. Check whether either value makes the lines parallel or intersecting, and describe the case k=1.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
For L₁:r=(1,0,0)+s(1,1,0) and L₂:r=(0,0,2)+t(2,2,0), explain why the skew-distance formula cannot be used. Find the shortest distance by another valid method.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
Locate the common perpendicular 公垂线
- Let $P=\mathbf a+s\mathbf u$ and $Q=\mathbf b+t\mathbf v$. The shortest connector obeys $(\mathbf q-\mathbf p)\cdot\mathbf u=0$ and $(\mathbf q-\mathbf p)\cdot\mathbf v=0$. Solve these two simultaneous equations for the parameters.
- Give the common perpendicular as a line through $P$ in direction $\mathbf u\times\mathbf v$, or through $Q$ in the same direction. Check it meets both original lines and that the segment length agrees with the distance formula. A perpendicular direction through an arbitrary point is insufficient.
For the two lines above,
Orthogonality gives $2s-t=3$ and $s-2t=4$, hence $s=2/3$ and $t=-5/3$. Therefore
is the common perpendicular. At $k=1/3$ it reaches $Q$, and $|PQ|=\sqrt3/3$ agrees with $1/\sqrt3$.
For L₁:r=s(1,2,1) and L₂:r=(2,0,1)+t(0,1,1), prove the lines are skew, find their shortest distance, locate the nearest points P and Q, and give the common-perpendicular line. Verify your answer.
State the geometric conditions, calculate with valid non-zero directions, and verify the final point, line or distance by substitution and orthogonality.
Check the geometric conditions
- A normal must be non-zero, a plane needs two independent directions, and the skew-distance denominator requires non-parallel lines. State parameter restrictions before squaring distance equations; absolute values can give more than one valid value.
- Substitute a proposed point into its plane, check an intersection line in both planes, and verify both orthogonality equations for a common perpendicular. These checks explain why the result solves the geometry.
Use non-zero directions and normals. A shortest connector must meet both objects and satisfy the required perpendicular conditions; a length alone does not locate it.
Two vectors have magnitudes 5 and 6 with angle 90°. What is the magnitude of their vector product?
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5 times 6 times 1 is 30. Other options — 6 (the lesson product for magnitudes 3 and 4 at 30 degrees); 11 (the sum of the magnitudes).