Question Statement Find a unit vector perpendicular to the plane containing a βΎ \underline{a} a β b βΎ \underline{b} b β
Given vectors:
(i) a βΎ = 2 i βΎ β 6 j βΎ β 3 k βΎ \underline{\mathbf{a}} = 2 \underline{\mathbf{i}} - 6 \underline{\mathbf{j}} - 3 \underline{\mathbf{k}} a β = 2 i β β 6 j β β 3 k β b βΎ = 4 i βΎ + 3 j βΎ β k βΎ \underline{\mathbf{b}} = 4 \underline{\mathbf{i}} + 3 \underline{\mathbf{j}} - \underline{\mathbf{k}} b β = 4 i β + 3 j β β k β
Background and Explanation In vector algebra, the cross product of two vectors a \mathbf{a} a b \mathbf{b} b perpendicular to the plane containing a \mathbf{a} a b \mathbf{b} b
To find the sine of the angle ΞΈ \theta ΞΈ
sin β‘ ΞΈ = β£ a Γ b β£ β£ a β£ β
β£ b β£ \sin \theta = \frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}| \cdot |\mathbf{b}|} sin ΞΈ = β£ a β£ β
β£ b β£ β£ a Γ b β£ β Where:
a Γ b \mathbf{a} \times \mathbf{b} a Γ b β£ a β£ |\mathbf{a}| β£ a β£ β£ b β£ |\mathbf{b}| β£ b β£ a \mathbf{a} a b \mathbf{b} b
Solution Part (i): Compute the unit vector perpendicular to the plane and the sine of the angle between a βΎ \underline{a} a β b βΎ \underline{b} b β
Given vectors:
a βΎ = 2 i βΎ β 6 j βΎ β 3 k βΎ , b βΎ = 4 i βΎ + 3 j βΎ β k βΎ \underline{\mathbf{a}} = 2\underline{\mathbf{i}} - 6\underline{\mathbf{j}} - 3\underline{\mathbf{k}}, \quad \underline{\mathbf{b}} = 4\underline{\mathbf{i}} + 3\underline{\mathbf{j}} - \underline{\mathbf{k}} a β = 2 i β β 6 j β β 3 k β , b β = 4 i β + 3 j β β k β First, we compute the cross product a βΎ Γ b βΎ \underline{a} \times \underline{b} a β Γ b β
a βΎ Γ b βΎ = β£ i j k 2 β 6 β 34 3 β 1 β£ \underline{a} \times \underline{b} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}
2 & -6 & -3
4 & 3 & -1
\end{vmatrix} a β Γ b β = β i β j β k 2 β β 6 β β 34 β 3 β β 1 β β Expanding the determinant:
a βΎ Γ b βΎ = i ( 6 + 9 ) β j ( β 2 + 12 ) + k ( 6 + 24 ) \underline{a} \times \underline{b} = \mathbf{i}(6 + 9) - \mathbf{j}(-2 + 12) + \mathbf{k}(6 + 24) a β Γ b β = i ( 6 + 9 ) β j ( β 2 + 12 ) + k ( 6 + 24 ) a βΎ Γ b βΎ = 15 i β 10 j + 30 k \underline{a} \times \underline{b} = 15 \mathbf{i} - 10 \mathbf{j} + 30 \mathbf{k} a β Γ b β = 15 i β 10 j + 30 k Next, compute the magnitude of the cross product:
β£ a βΎ Γ b βΎ β£ = ( 15 ) 2 + ( β 10 ) 2 + ( 30 ) 2 = 1225 = 35 |\underline{a} \times \underline{b}| = \sqrt{(15)^2 + (-10)^2 + (30)^2} = \sqrt{1225} = 35 β£ a β Γ b β β£ = ( 15 ) 2 + ( β 10 ) 2 + ( 30 ) 2 β = 1225 β = 35 The unit vector perpendicular to the plane is:
n ^ = a βΎ Γ b βΎ β£ a βΎ Γ b βΎ β£ = 15 i β 10 j + 30 k 35 = 3 7 i β 2 7 j + 6 7 k \hat{n} = \frac{\underline{a} \times \underline{b}}{|\underline{a} \times \underline{b}|} = \frac{15\mathbf{i} - 10\mathbf{j} + 30\mathbf{k}}{35} = \frac{3}{7}\mathbf{i} - \frac{2}{7}\mathbf{j} + \frac{6}{7}\mathbf{k} n ^ = β£ a β Γ b β β£ a β Γ b β β = 35 15 i β 10 j + 30 k β = 7 3 β i β 7 2 β j + 7 6 β k Next, compute the magnitudes of a βΎ \underline{a} a β b βΎ \underline{b} b β
β£ a βΎ β£ = 2 2 + ( β 6 ) 2 + ( β 3 ) 2 = 4 + 36 + 9 = 7 |\underline{a}| = \sqrt{2^2 + (-6)^2 + (-3)^2} = \sqrt{4 + 36 + 9} = 7 β£ a β β£ = 2 2 + ( β 6 ) 2 + ( β 3 ) 2 β = 4 + 36 + 9 β = 7 β£ b βΎ β£ = 4 2 + 3 2 + ( β 1 ) 2 = 16 + 9 + 1 = 26 |\underline{b}| = \sqrt{4^2 + 3^2 + (-1)^2} = \sqrt{16 + 9 + 1} = \sqrt{26} β£ b β β£ = 4 2 + 3 2 + ( β 1 ) 2 β = 16 + 9 + 1 β = 26 β Now, calculate the sine of the angle ΞΈ \theta ΞΈ a βΎ \underline{a} a β b βΎ \underline{b} b β
sin β‘ ΞΈ = β£ a βΎ Γ b βΎ β£ β£ a βΎ β£ β
β£ b βΎ β£ = 35 7 β
26 = 5 26 \sin \theta = \frac{|\underline{a} \times \underline{b}|}{|\underline{a}| \cdot |\underline{b}|} = \frac{35}{7 \cdot \sqrt{26}} = \frac{5}{\sqrt{26}} sin ΞΈ = β£ a β β£ β
β£ b β β£ β£ a β Γ b β β£ β = 7 β
26 β 35 β = 26 β 5 β Part (ii): Compute the unit vector perpendicular to the plane and the sine of the angle between a βΎ = β i β j β k \underline{a} = -\mathbf{i} - \mathbf{j} - \mathbf{k} a β = β i β j β k b βΎ = 2 i β 3 j + 4 k \underline{b} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} b β = 2 i β 3 j + 4 k
Given vectors:
a βΎ = β i β j β k , b βΎ = 2 i β 3 j + 4 k \underline{\mathbf{a}} = -\mathbf{i} - \mathbf{j} - \mathbf{k}, \quad \underline{\mathbf{b}} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} a β = β i β j β k , b β = 2 i β 3 j + 4 k First, compute the cross product a βΎ Γ b βΎ \underline{a} \times \underline{b} a β Γ b β
a βΎ Γ b βΎ = β£ i j k β 1 β 1 β 12 β 3 4 β£ \underline{a} \times \underline{b} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}
-1 & -1 & -1
2 & -3 & 4
\end{vmatrix} a β Γ b β = β i β j β k β 1 β β 1 β β 12 β β 3 β 4 β β Expanding the determinant:
a βΎ Γ b βΎ = i ( β 4 β 3 ) β j ( β 4 + 2 ) + k ( 3 + 2 ) \underline{a} \times \underline{b} = \mathbf{i}(-4 - 3) - \mathbf{j}(-4 + 2) + \mathbf{k}(3 + 2) a β Γ b β = i ( β 4 β 3 ) β j ( β 4 + 2 ) + k ( 3 + 2 ) a βΎ Γ b βΎ = β 7 i + 2 j + 5 k \underline{a} \times \underline{b} = -7\mathbf{i} + 2\mathbf{j} + 5\mathbf{k} a β Γ b β = β 7 i + 2 j + 5 k Now, compute the magnitude of the cross product:
β£ a βΎ Γ b βΎ β£ = ( β 7 ) 2 + 2 2 + 5 2 = 49 + 4 + 25 = 78 |\underline{a} \times \underline{b}| = \sqrt{(-7)^2 + 2^2 + 5^2} = \sqrt{49 + 4 + 25} = \sqrt{78} β£ a β Γ b β β£ = ( β 7 ) 2 + 2 2 + 5 2 β = 49 + 4 + 25 β = 78 β Next, compute the magnitudes of a βΎ \underline{a} a β b βΎ \underline{b} b β
β£ a βΎ β£ = ( β 1 ) 2 + ( β 1 ) 2 + ( β 1 ) 2 = 3 |\underline{a}| = \sqrt{(-1)^2 + (-1)^2 + (-1)^2} = \sqrt{3} β£ a β β£ = ( β 1 ) 2 + ( β 1 ) 2 + ( β 1 ) 2 β = 3 β β£ b βΎ β£ = 2 2 + ( β 3 ) 2 + 4 2 = 29 |\underline{b}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{29} β£ b β β£ = 2 2 + ( β 3 ) 2 + 4 2 β = 29 β Now, calculate the sine of the angle ΞΈ \theta ΞΈ a βΎ \underline{a} a β b βΎ \underline{b} b β
sin β‘ ΞΈ = β£ a βΎ Γ b βΎ β£ β£ a βΎ β£ β
β£ b βΎ β£ = 78 3 β
29 = 78 87 = 26 29 \sin \theta = \frac{|\underline{a} \times \underline{b}|}{|\underline{a}| \cdot |\underline{b}|} = \frac{\sqrt{78}}{\sqrt{3} \cdot \sqrt{29}} = \sqrt{\frac{78}{87}} = \sqrt{\frac{26}{29}} sin ΞΈ = β£ a β β£ β
β£ b β β£ β£ a β Γ b β β£ β = 3 β β
29 β 78 β β = 87 78 β β = 29 26 β β Part (iii): Compute the unit vector perpendicular to the plane and the sine of the angle between a βΎ = 2 i β 2 j + 4 k \underline{a} = 2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} a β = 2 i β 2 j + 4 k b βΎ = β i + 3 j β 2 k \underline{b} = -\mathbf{i} + 3\mathbf{j} - 2\mathbf{k} b β = β i + 3 j β 2 k
Given vectors:
a βΎ = 2 i β 2 j + 4 k , b βΎ = β i + 3 j β 2 k \underline{\mathbf{a}} = 2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}, \quad \underline{\mathbf{b}} = -\mathbf{i} + 3\mathbf{j} - 2\mathbf{k} a β = 2 i β 2 j + 4 k , b β = β i + 3 j β 2 k Compute the cross product a βΎ Γ b βΎ \underline{a} \times \underline{b} a β Γ b β
a βΎ Γ b βΎ = β£ i j k 2 β 2 4 β 1 1 β 2 β£ \underline{a} \times \underline{b} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}
2 & -2 & 4
-1 & 1 & -2
\end{vmatrix} a β Γ b β = β i β j β k 2 β β 2 β 4 β 1 β 1 β β 2 β β Expanding the determinant:
a βΎ Γ b βΎ = i ( 4 β 4 ) β j ( β 4 + 4 ) + k ( 2 β 2 ) \underline{a} \times \underline{b} = \mathbf{i}(4 - 4) - \mathbf{j}(-4 + 4) + \mathbf{k}(2 - 2) a β Γ b β = i ( 4 β 4 ) β j ( β 4 + 4 ) + k ( 2 β 2 ) a βΎ Γ b βΎ = 0 i β 0 j + 0 k \underline{a} \times \underline{b} = 0 \mathbf{i} - 0 \mathbf{j} + 0 \mathbf{k} a β Γ b β = 0 i β 0 j + 0 k Since the cross product is the null vector , the vectors a βΎ \underline{a} a β b βΎ \underline{b} b β parallel and the sine of the angle between them is:
sin β‘ ΞΈ = 0 \sin \theta = 0 sin ΞΈ = 0 Part (iv): Compute the unit vector perpendicular to the plane and the sine of the angle between a βΎ = i + j \underline{a} = \mathbf{i} + \mathbf{j} a β = i + j b βΎ = i β j \underline{b} = \mathbf{i} - \mathbf{j} b β = i β j
Given vectors:
a βΎ = i + j , b βΎ = i β j \underline{\mathbf{a}} = \mathbf{i} + \mathbf{j}, \quad \underline{\mathbf{b}} = \mathbf{i} - \mathbf{j} a β = i + j , b β = i β j Compute the cross product a βΎ Γ b βΎ \underline{a} \times \underline{b} a β Γ b β
a βΎ Γ b βΎ = β£ i j k 1 1 01 β 1 0 β£ \underline{a} \times \underline{b} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}
1 & 1 & 0
1 & -1 & 0
\end{vmatrix} a β Γ b β = β i β j β k 1 β 1 β 01 β β 1 β 0 β β Expanding the determinant:
a βΎ Γ b βΎ = i ( 0 β 0 ) β j ( 0 β 0 ) + k ( β 1 β 1 ) \underline{a} \times \underline{b} = \mathbf{i}(0 - 0) - \mathbf{j}(0 - 0) + \mathbf{k}(-1 - 1) a β Γ b β = i ( 0 β 0 ) β j ( 0 β 0 ) + k ( β 1 β 1 ) a βΎ Γ b βΎ = β 2 k \underline{a} \times \underline{b} = -2 \mathbf{k} a β Γ b β = β 2 k Now, compute the magnitude of the cross product:
β£ a βΎ Γ b βΎ β£ = ( β 2 ) 2 = 4 = 2 |\underline{a} \times \underline{b}| = \sqrt{(-2)^2} = \sqrt{4} = 2 β£ a β Γ b β β£ = ( β 2 ) 2 β = 4 β = 2 Now, compute the magnitudes of a βΎ \underline{a} a β b βΎ \underline{b} b β
β£ a βΎ β£ = 1 2 + 1 2 = 2 , β£ b βΎ β£ = 1 2 + ( β 1 ) 2 = 2 |\underline{a}| = \sqrt{1^2 + 1^2} = \sqrt{2}, \quad |\underline{b}| = \sqrt{1^2 + (-1)^2} = \sqrt{2} β£ a β β£ = 1 2 + 1 2 β = 2 β , β£ b β β£ = 1 2 + ( β 1 ) 2 β = 2 β Now, calculate the sine of the angle ΞΈ \theta ΞΈ a βΎ \underline{a} a β b βΎ \underline{b} b β
sin β‘ ΞΈ = β£ a βΎ Γ b βΎ β£ β£ a βΎ β£ β
β£ b βΎ β£ = 2 2 Γ 2 = 2 2 = 1 \sin \theta = \frac{|\underline{a} \times \underline{b}|}{|\underline{a}| \cdot |\underline{b}|} = \frac{2}{\sqrt{2} \times \sqrt{2}} = \frac{2}{2} = 1 sin ΞΈ = β£ a β β£ β
β£ b β β£ β£ a β Γ b β β£ β = 2 β Γ 2 β 2 β = 2 2 β = 1
Cross Product :
a Γ b = β£ i j k a 1 a 2 a 3 b 1 b 2 b 3 β£ \mathbf{a} \times \mathbf{b} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}
a_1 & a_2 & a_3
b_1 & b_2 & b_3
\end{vmatrix} a Γ b = β i β j β k a 1 β β a 2 β β a 3 β b 1 β β b 2 β β b 3 β β β
Perpendicular Unit Vector :
n ^ = a Γ b β£ a Γ b β£ \hat{n} = \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|} n ^ = β£ a Γ b β£ a Γ b β
Sine of the Angle :
sin β‘ ΞΈ = β£ a Γ b β£ β£ a β£ β
β£ b β£ \sin \theta = \frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}| \cdot |\mathbf{b}|} sin ΞΈ = β£ a β£ β
β£ b β£ β£ a Γ b β£ β Summary of Steps
Compute the cross product a βΎ Γ b βΎ \underline{a} \times \underline{b} a β Γ b β
Compute the magnitude of the cross product and normalize it to find the unit vector.
Compute the magnitudes of a βΎ \underline{a} a β b βΎ \underline{b} b β
Use the formula for sin β‘ ΞΈ \sin \theta sin ΞΈ
