• <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(836) 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

            91久久国产视频| 久久亚洲日韩看片无码| 精品国产乱码久久久久久呢| 欧美激情精品久久久久久久| 亚洲午夜无码久久久久小说| 久久受www免费人成_看片中文 | 国产高清美女一级a毛片久久w| 国产午夜精品理论片久久| 久久综合久久鬼色| 久久久久久九九99精品| 伊人久久综合热线大杳蕉下载| 91精品国产91热久久久久福利| 精品久久久久久国产免费了| 亚洲色欲久久久综合网东京热| 亚洲αv久久久噜噜噜噜噜| 久久综合给合久久国产免费| 91久久福利国产成人精品| 久久人人爽人人爽人人片AV东京热| 久久超乳爆乳中文字幕| 少妇无套内谢久久久久| 成人国内精品久久久久一区| 亚洲国产精品狼友中文久久久 | 影音先锋女人AV鲁色资源网久久| 久久一区二区三区99| 午夜不卡888久久| 国产成人久久777777| 国产成人香蕉久久久久| 久久精品无码一区二区日韩AV| 久久99精品久久久久婷婷| 97久久香蕉国产线看观看| 999久久久国产精品| 91精品国产91久久| 久久99国产精品久久99小说| 久久综合色老色| 久久精品国产99国产电影网 | 亚洲综合婷婷久久| 久久免费看黄a级毛片| 久久久久久亚洲精品成人| 久久久无码精品午夜| 综合久久一区二区三区| 麻豆亚洲AV永久无码精品久久|