Given that A → = 2 i → + 3 j → − k → \overrightarrow{A} = 2\overrightarrow{i} + 3\overrightarrow{j} - \overrightarrow{k} A = 2 i + 3 j − k B → = i → − j → + 2 k → \overrightarrow{B} = \overrightarrow{i} -\overrightarrow{j} +2 \overrightarrow{k} B = i − j + 2 k C → = 3 i → + 4 j → + k → \overrightarrow{C} = 3\overrightarrow{i} + 4\overrightarrow{j}+\overrightarrow{k} C = 3 i + 4 j + k
a)find the unit vector along the direction of nA, for n<0
e → = − A → ∣ A → ∣ = − 2 i → + 3 j → − k → 2 2 + 3 2 + ( − 1 ) 2 = \overrightarrow{e}=-\frac{\overrightarrow{A}}{|\overrightarrow{A}|}=-\frac{2\overrightarrow{i} + 3\overrightarrow{j} - \overrightarrow{k}}{\sqrt{2^2+3^2+(-1)^2}}= e = − ∣ A ∣ A = − 2 2 + 3 2 + ( − 1 ) 2 2 i + 3 j − k = − 2 14 i → − 3 14 j → + 1 14 k → -\frac{2}{\sqrt{14}}\overrightarrow{i} -\frac{3}{\sqrt{14}}\overrightarrow{j} +\frac{1}{\sqrt{14}} \overrightarrow{k} − 14 2 i − 14 3 j + 14 1 k
b) find the length of the projection of vector A on vector C
p r o j C → A → = A → ⋅ C → ∣ C → ∣ = proj_{\overrightarrow{C}}\overrightarrow{A}=\frac{\overrightarrow{A}\cdot\overrightarrow{C}}{|\overrightarrow{C}|}= p ro j C A = ∣ C ∣ A ⋅ C = 2 ⋅ 3 + 3 ⋅ 4 + ( − 1 ) ⋅ 1 3 2 + 4 2 + 1 2 = 17 26 \frac{2\cdot3+3\cdot4+(-1)\cdot1}{\sqrt{3^2+4^2+1^2}}=\frac{17}{\sqrt{26}} 3 2 + 4 2 + 1 2 2 ⋅ 3 + 3 ⋅ 4 + ( − 1 ) ⋅ 1 = 26 17
c)find the length of projection of vector C and vector A
p r o j A → C → = A → ⋅ C → ∣ A → ∣ = proj_{\overrightarrow{A}}\overrightarrow{C}=\frac{\overrightarrow{A}\cdot\overrightarrow{C}}{|\overrightarrow{A}|}= p ro j A C = ∣ A ∣ A ⋅ C = 2 ⋅ 3 + 3 ⋅ 4 + ( − 1 ) ⋅ 1 2 2 + 3 2 + ( − 1 ) 2 = 17 14 \frac{2\cdot3+3\cdot4+(-1)\cdot1}{\sqrt{2^2+3^2+(-1)^2}}=\frac{17}{\sqrt{14}} 2 2 + 3 2 + ( − 1 ) 2 2 ⋅ 3 + 3 ⋅ 4 + ( − 1 ) ⋅ 1 = 14 17
d) find the sine of the angle between vector A and vector C
∣ A → × C → ∣ = ∣ A → ∣ ⋅ ∣ C → ∣ ⋅ sin α |\overrightarrow{A}\times\overrightarrow{C}|=|\overrightarrow{A}|\cdot|\overrightarrow{C}|\cdot\sin{\alpha} ∣ A × C ∣ = ∣ A ∣ ⋅ ∣ C ∣ ⋅ sin α α \alpha α A → \overrightarrow{A} A C → \overrightarrow{C} C
sin α = ∣ A → × C → ∣ ∣ A → ∣ ⋅ ∣ C → ∣ \sin{\alpha}=\frac{|\overrightarrow{A}\times\overrightarrow{C}|}{|\overrightarrow{A}|\cdot|\overrightarrow{C}|} sin α = ∣ A ∣ ⋅ ∣ C ∣ ∣ A × C ∣
A → × C → = ∣ i → j → k → 2 3 − 1 3 4 1 ∣ = \overrightarrow{A}\times\overrightarrow{C}=\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\
2 & 3 & -1 \\
3 & 4 & 1
\end{vmatrix}= A × C = ∣ ∣ i 2 3 j 3 4 k − 1 1 ∣ ∣ = i → ( 3 ⋅ 1 − 4 ⋅ ( − 1 ) ) − j → ( 2 ⋅ 1 − 3 ⋅ ( − 1 ) ) + \overrightarrow{i}(3\cdot1-4\cdot(-1))-\overrightarrow{j}(2\cdot1-3\cdot(-1))+ i ( 3 ⋅ 1 − 4 ⋅ ( − 1 )) − j ( 2 ⋅ 1 − 3 ⋅ ( − 1 )) + k → ( 2 ⋅ 4 − 3 ⋅ 3 ) = \overrightarrow{k}(2\cdot4-3\cdot3)= k ( 2 ⋅ 4 − 3 ⋅ 3 ) = 7 i → − 5 j → − k → 7\overrightarrow{i}-5\overrightarrow{j}-\overrightarrow{k} 7 i − 5 j − k
sin α = 7 2 + ( − 5 ) 2 + ( − 1 ) 2 2 2 + 3 2 + ( − 1 ) 2 ⋅ 3 2 + 4 2 + 1 2 = 75 14 ⋅ 26 = 75 364 = \sin{\alpha}=\frac{\sqrt{7^2+(-5)^2+(-1)^2}}{\sqrt{2^2+3^2+(-1)^2}\cdot\sqrt{3^2+4^2+1^2}}=\frac{\sqrt{75}}{\sqrt{14}\cdot\sqrt{26}}=\frac{\sqrt{75}}{\sqrt{364}}= sin α = 2 2 + 3 2 + ( − 1 ) 2 ⋅ 3 2 + 4 2 + 1 2 7 2 + ( − 5 ) 2 + ( − 1 ) 2 = 14 ⋅ 26 75 = 364 75 = 5 2 3 91 \frac 52\sqrt{\frac{3}{91}} 2 5 91 3
#340153 on Dec 2023
Comments