標題:
vector
發問:
I know part (a) of both the questions only, cannot find out the answers of Part (b) so please help, thank you~~ picture: http://upload.lsforum.net/users/public/f384951-11w64.png 更新: and this too, i only konw part (a) 3. http://upload.lsforum.net/users/public/c207111-11-2c222.png 更新 2: 我計到第三題了,所以不用回答了,謝謝!!! 更新 3: 請問有沒有人幫忙?? 更新 4: Ok and thank you http://upload.lsforum.net/users/public/b525253-11b64.png 更新 5: 我知我不明白的地方了... 我以為計cos∠DEC,要ED．EC, 原來OD．BC 都得??!! 但OD比ED長, BC比EC長 ,, OD．BC 不是"痴"住∠DEC,你明我的意思嗎? 更新 6: How about this ??thanks for your time!!!! http://upload.lsforum.net/users/public/l4008911-3-2w222.png 更新 7: 明白,謝謝!
最佳解答:
Since you have done part (a) for both questions, it is better for you to give the answers to (a) because I need to use those result to do (b). If you provide more information, it facilitates our thoughts and you can get your answers from us faster... 2013-11-03 17:43:05 補充： 1 (b) From (a), you have OD = a + 0.5 c and OE = c + 0.5a Let θ be your required ∠DOE. OD．OE = (a + 0.5 c)．(c + 0.5a) = 0.5|a|2 + 0.5|c|2 = |a|2 because |a| = |c| [Note that a．c = 0 since they are perpendicular.] Also, OD．OE = |OD||OE| cos θ 2013-11-03 17:45:09 補充： Note that |OD| = |OE| = √(OD．OD) = √[(a + 0.5c)．(a + 0.5c)] = √(|a|2 + 0.25|c|2) = √(1.25|a|2) = √1.25 |a| Therefore, |a|2 = √1.25 |a| √1.25 |a| cos θ |a|2 = 1.25 |a|2 cos θ cos θ = 4/5 θ = 36.86989765° 2013-11-03 17:48:12 補充： 你提供 (a) 的答案，我再做 2 (b) 從你這幾天的發問可以看出你不太熟悉 dot product 的應用。 努力呀！ 記住，一方面 dot product 可以有角度的計算 a．b = |a| |b| cos (included angle) 另一方面， dot product （如有i, j, k）可計出數值～ (xi + yj)．(ai + bj) = xa + yb 這就可能可以設立方程讓你計算。 2013-11-03 18:12:04 補充： 1 (b) From (a), you have OD = a + 0.5 c and OE = c + 0.5a Let θ be your required ∠DOE. OD．OE = (a + 0.5 c)．(c + 0.5a) = 0.5|a|2 + 0.5|c|2 = |a|2 because |a| = |c| [Note that a．c = 0 since they are perpendicular.] Also, OD．OE = |OD||OE| cos θ Note that |OD| = |OE| = √(OD．OD) = √[(a + 0.5c)．(a + 0.5c)] = √(|a|2 + 0.25|c|2) = √(1.25|a|2) = √1.25 |a| Therefore, |a|2 = √1.25 |a| √1.25 |a| cos θ |a|2 = 1.25 |a|2 cos θ cos θ = 4/5 θ = 36.86989765° ≈ 37° ================================================== 2 (b) From (a), you have OD = (1/3)a + (2/3)b and BC = (1/2)a - b a．b = |a| |b| cos60° = 3*2*(1/2) = 3 For calculating ∠DEC, consider OD．BC = |OD| |BC| cos∠DEC |OD|2 = OD．OD = [(1/3)a + (2/3)b]．[(1/3)a + (2/3)b] = (1/9)(9) + (4/9)(4) + 2(1/3)(2/3)a．b = 1+ 16/9 + 4/9*3 = 37/9 |OD| = √37 / 3 |BC|2 = BC．BC = [(1/2)a - b]．[(1/2)a - b] = (1/4)(9) + 4 - 2(1/2)a．b = 9/4 + 4 - 3 = 13/4 |BC| = √13 / 2 Therefore, OD．BC = |OD| |BC| cos∠DEC OD．BC = (√37 / 3) (√13 / 2) cos∠DEC On the other hand, OD．BC = [(1/3)a + (2/3)b]．[(1/2)a - b] = (1/6)(9) - (2/3)(4) = -7/6 With (√37 / 3) (√13 / 2) cos∠DEC = -7/6 cos∠DEC = -7/√481 ∠DEC = 108.6128902° ≈ 109° 〔由於時間關係，我沒有再檢查答案，但方法是正確的，請再自己檢查一次。〕 ================================================== 從你這幾天的發問可以看出你不太熟悉 dot product 的應用。 努力呀！ 記住，一方面 dot product 可以有角度的計算 a．b = |a| |b| cos (included angle) 另一方面， dot product （如有i, j, k）可計出數值～ (xi + yj)．(ai + bj) = xa + yb 這就可能可以設立方程讓你計算。 2013-11-03 19:33:32 補充： 我明你的意思～ 但 vector 是可以移位的，而且重點只是夾角～ ED和EC的夾角 = OD和BC的夾角～ 2013-11-03 22:18:01 補充： 回應你於2013-11-03 21:52:58 的補充發問： 對，因為是同一條直線的部份。 對於現在的課題，即是你要證明 dot product = 0
vector
發問:
I know part (a) of both the questions only, cannot find out the answers of Part (b) so please help, thank you~~ picture: http://upload.lsforum.net/users/public/f384951-11w64.png 更新: and this too, i only konw part (a) 3. http://upload.lsforum.net/users/public/c207111-11-2c222.png 更新 2: 我計到第三題了,所以不用回答了,謝謝!!! 更新 3: 請問有沒有人幫忙?? 更新 4: Ok and thank you http://upload.lsforum.net/users/public/b525253-11b64.png 更新 5: 我知我不明白的地方了... 我以為計cos∠DEC,要ED．EC, 原來OD．BC 都得??!! 但OD比ED長, BC比EC長 ,, OD．BC 不是"痴"住∠DEC,你明我的意思嗎? 更新 6: How about this ??thanks for your time!!!! http://upload.lsforum.net/users/public/l4008911-3-2w222.png 更新 7: 明白,謝謝!
最佳解答:
Since you have done part (a) for both questions, it is better for you to give the answers to (a) because I need to use those result to do (b). If you provide more information, it facilitates our thoughts and you can get your answers from us faster... 2013-11-03 17:43:05 補充： 1 (b) From (a), you have OD = a + 0.5 c and OE = c + 0.5a Let θ be your required ∠DOE. OD．OE = (a + 0.5 c)．(c + 0.5a) = 0.5|a|2 + 0.5|c|2 = |a|2 because |a| = |c| [Note that a．c = 0 since they are perpendicular.] Also, OD．OE = |OD||OE| cos θ 2013-11-03 17:45:09 補充： Note that |OD| = |OE| = √(OD．OD) = √[(a + 0.5c)．(a + 0.5c)] = √(|a|2 + 0.25|c|2) = √(1.25|a|2) = √1.25 |a| Therefore, |a|2 = √1.25 |a| √1.25 |a| cos θ |a|2 = 1.25 |a|2 cos θ cos θ = 4/5 θ = 36.86989765° 2013-11-03 17:48:12 補充： 你提供 (a) 的答案，我再做 2 (b) 從你這幾天的發問可以看出你不太熟悉 dot product 的應用。 努力呀！ 記住，一方面 dot product 可以有角度的計算 a．b = |a| |b| cos (included angle) 另一方面， dot product （如有i, j, k）可計出數值～ (xi + yj)．(ai + bj) = xa + yb 這就可能可以設立方程讓你計算。 2013-11-03 18:12:04 補充： 1 (b) From (a), you have OD = a + 0.5 c and OE = c + 0.5a Let θ be your required ∠DOE. OD．OE = (a + 0.5 c)．(c + 0.5a) = 0.5|a|2 + 0.5|c|2 = |a|2 because |a| = |c| [Note that a．c = 0 since they are perpendicular.] Also, OD．OE = |OD||OE| cos θ Note that |OD| = |OE| = √(OD．OD) = √[(a + 0.5c)．(a + 0.5c)] = √(|a|2 + 0.25|c|2) = √(1.25|a|2) = √1.25 |a| Therefore, |a|2 = √1.25 |a| √1.25 |a| cos θ |a|2 = 1.25 |a|2 cos θ cos θ = 4/5 θ = 36.86989765° ≈ 37° ================================================== 2 (b) From (a), you have OD = (1/3)a + (2/3)b and BC = (1/2)a - b a．b = |a| |b| cos60° = 3*2*(1/2) = 3 For calculating ∠DEC, consider OD．BC = |OD| |BC| cos∠DEC |OD|2 = OD．OD = [(1/3)a + (2/3)b]．[(1/3)a + (2/3)b] = (1/9)(9) + (4/9)(4) + 2(1/3)(2/3)a．b = 1+ 16/9 + 4/9*3 = 37/9 |OD| = √37 / 3 |BC|2 = BC．BC = [(1/2)a - b]．[(1/2)a - b] = (1/4)(9) + 4 - 2(1/2)a．b = 9/4 + 4 - 3 = 13/4 |BC| = √13 / 2 Therefore, OD．BC = |OD| |BC| cos∠DEC OD．BC = (√37 / 3) (√13 / 2) cos∠DEC On the other hand, OD．BC = [(1/3)a + (2/3)b]．[(1/2)a - b] = (1/6)(9) - (2/3)(4) = -7/6 With (√37 / 3) (√13 / 2) cos∠DEC = -7/6 cos∠DEC = -7/√481 ∠DEC = 108.6128902° ≈ 109° 〔由於時間關係，我沒有再檢查答案，但方法是正確的，請再自己檢查一次。〕 ================================================== 從你這幾天的發問可以看出你不太熟悉 dot product 的應用。 努力呀！ 記住，一方面 dot product 可以有角度的計算 a．b = |a| |b| cos (included angle) 另一方面， dot product （如有i, j, k）可計出數值～ (xi + yj)．(ai + bj) = xa + yb 這就可能可以設立方程讓你計算。 2013-11-03 19:33:32 補充： 我明你的意思～ 但 vector 是可以移位的，而且重點只是夾角～ ED和EC的夾角 = OD和BC的夾角～ 2013-11-03 22:18:01 補充： 回應你於2013-11-03 21:52:58 的補充發問： 對，因為是同一條直線的部份。 對於現在的課題，即是你要證明 dot product = 0
此文章來自奇摩知識+如有不便請留言告知
其他解答:
文章標籤
全站熱搜
留言列表
