• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>

            天行健 君子當自強而不息

            D3D編程必備的數學知識(2)

            向量相加

            我們能夠通過分別把兩個向量的各個分量相加得到向量之和,注意在相加之前必須保證它們有相同的維數。

            u + v = (ux+ vx, uy+ vy, uz+ vz)

             

            圖5顯示的是幾何學上的向量相加。

            兩個向量相加的代碼,我們使用重載的加法操作符:

            D3DXVECTOR3 u(2.0f, 0.0f, 1.0f);

            D3DXVECTOR3 v(0.0f, -1.0f, 5.0f);

            // (2.0 + 0.0,  0.0 + (-1.0),  1.0 + 5.0)

            D3DXVECTOR3 sum = u + v; // = (2.0f, -1.0f, 6.0f)

             

             

             

            向量相減

            和加法類似,通過分別把兩個向量的各個分量相減得到向量之差。再次重聲兩個向量必須是相同維數。

            u-v = u + (-v) = (ux - vx, uy - vy, uz - vz)

             

            圖6顯示的是幾何學上的向量相減。

             

            兩個向量相減的代碼,我們使用重載的減法操作符:

            D3DXVECTOR3 u(2.0f, 0.0f, 1.0f);

            D3DXVECTOR3 v(0.0f, -1.0f, 5.0f);

            D3DXVECTOR3 difference = u - v; // = (2.0f, 1.0f, -4.0f)

            圖6顯示,向量減法得到一個從v向量終點到u向量終點的向量。假如我們解釋uv的分量,我們能用向量相減找到從一個點到另一個點的向量。這是非常方便的操作,因為我們常常想找到從一個點到另一個點的方向向量。

             

             

            標量與向量的乘積

            我們能用一個標量與向量相乘,就象名字暗示的一樣,向量按比例變化。這種運算不會改變向量的方向,除非標量是負數,這種情況向量方向相反。

            ku = (kux, kuy, kuz)

            D3DXVECTOR3類提供了向量與標量乘法的操作符。

            D3DXVECTOR3 u(1.0f, 1.0f, -1.0f);

            D3DXVECTOR3 scaledVec = u * 10.0f; // = (10.0f, 10.0f, -10.0f)

             

            點積

            數學上定義點積是兩個向量的乘積。按下面等式計算:

             

            u.v = uxvx + uyvy + uzvz = s

            The above formula does not present an obvious geometric meaning. Using the law of cosines, we can find the relationship u.v = ∥u∥∥v∥ cosθ , which says that the dot product between two vectors is the cosine of the angle between them scaled by the vectors' magnitudes. Thus, if both u and v are unit vectors, then u.v is the cosine of the angle between them.

            Some useful properties of the dot product:

            • If u.v = 0, then uv.

            • If u.v > 0, then the angle θ, between the two vectors is less than 90 degrees.

            • If u.v < 0, then the angle θ, between the two vectors is greater than 90 degrees.

              Note 

            The ⊥ symbol means "orthogonal," which is synonymous with the term "perpendicular."

            We use the following D3DX function to compute the dot product between two vectors:

            FLOAT D3DXVec3Dot(          // Returns the result.
            CONST D3DXVECTOR3* pV1, // Left sided operand.
            CONST D3DXVECTOR3* pV2 // Right sided operand.
            );

            D3DXVECTOR3 u(1.0f, -1.0f, 0.0f);
            D3DXVECTOR3 v(3.0f, 2.0f, 1.0f);

            // 1.0*3.0 + -1.0*2.0 + 0.0*1.0
            // = 3.0 + -2.0
            float dot = D3DXVec3Dot( &u, &v ); // = 1.0

            叉積

            第二種乘法在向量數學中叫叉積。不象點積,結果值是一個標量,叉積的結果值是另一個向量。通過把兩個向量uv相乘得到另一的向量p,向量p垂直于uv。也就是說向量p垂直于u并且垂直于u

            The cross product is computed like so:

            p = u×v = [(uyvz - uzvy), (uzvx - uxvz), (uxvy - uyvx)]

            In component form:

            px = (uyvz - uzvy)

            py = (uzvx - uxvz)

            pz = (uxvy - uyvx)

            Example: Find j = k × i = (0, 0, 1) × (1, 0, 0) and verify that j is orthogonal to both k and i.

            Solution:

            jx =(0(0)-1(0)) = 0

            jy =(1(1)-0(0) = 1

            jz=(0(0)-0(1) = 0

            So, j = (0, 1, 0). Recall from the section titled "Dot Products" that if u.v = 0, then uv Since j.k = 0 and j.i = 0, we know j is orthogonal to both k and i.

            We use the following D3DX function to compute the cross product between two vectors:

            D3DXVECTOR3 *D3DXVec3Cross(
            D3DXVECTOR3* pOut, // Result.
            CONST D3DXVECTOR3* pV1, // Left sided operand.
            CONST D3DXVECTOR3* pV2 // Right sided operand.
            );

            It is obvious from Figure 7 that the vector -p is also mutually orthogonal to both u and v. The order in which we perform the cross product determines whether we get p or -p as a result. In other words, u × v = -(v × u). This shows that the cross product is not commutative. You can determine the vector returned by the cross product by the left hand thumb rule. (We use a left hand rule because we are using a left-handed coordinate system. We would switch to the right hand rule if we were using a right-handed coordinate system.) If you curve the fingers of your left hand in the direction of the first vector toward the second vector, your thumb points in the direction of the returned vector.

             


            posted on 2008-03-12 10:58 lovedday 閱讀(825) 評論(0)  編輯 收藏 引用

            公告

            導航

            統計

            常用鏈接

            隨筆分類(178)

            3D游戲編程相關鏈接

            搜索

            最新評論

            亚洲女久久久噜噜噜熟女| 国产精品国色综合久久| 国产精品成人久久久久久久| 国产高清美女一级a毛片久久w| 久久人人爽人人澡人人高潮AV| 中文国产成人精品久久亚洲精品AⅤ无码精品 | 国产福利电影一区二区三区,免费久久久久久久精| 国产精品久久久久久影院 | 热综合一本伊人久久精品| 99久久香蕉国产线看观香| 国产精品岛国久久久久| 欧美日韩精品久久久久| 久久久久国产精品熟女影院 | 久久久高清免费视频| 久久精品国产久精国产| 中文字幕人妻色偷偷久久| 草草久久久无码国产专区| 麻豆亚洲AV永久无码精品久久| 久久精品国产亚洲7777| 2022年国产精品久久久久| 久久99精品国产麻豆宅宅| 亚洲欧美精品一区久久中文字幕 | 日韩久久久久久中文人妻| 久久久午夜精品| 性做久久久久久免费观看| 伊人丁香狠狠色综合久久| 国产精品久久成人影院| 久久久无码人妻精品无码| 亚洲精品综合久久| 中文字幕无码久久人妻| 综合久久精品色| 要久久爱在线免费观看| 一级女性全黄久久生活片免费| 久久久久99精品成人片牛牛影视| 国产精品一区二区久久精品无码| 精品亚洲综合久久中文字幕| 日产精品久久久一区二区| 久久99精品久久只有精品| 国产精品一久久香蕉国产线看| 久久青青草原亚洲av无码app| 狠狠色丁香婷婷综合久久来|