(1) If \( \vec{a} = \hat{i} + \hat{j}, \ \vec{b} = \hat{i} - \hat{j}, \ \vec{c} = 5\hat{i} + 2\hat{j} + 3\hat{k} \), find the value of \([ \vec{b}\ \vec{c}\ \vec{a} ]\).
We know that the scalar triple product
\([ \vec{b}\ \vec{c}\ \vec{a} ] = \begin{vmatrix} b_x & b_y & b_z \\ c_x & c_y & c_z \\ a_x & a_y & a_z \end{vmatrix}\)
Substituting the values:
\([ \vec{b}\ \vec{c}\ \vec{a} ] = \begin{vmatrix} 1 & -1 & 0\\ 5 & 2 & 3\\ 1 & 1 & 0 \end{vmatrix}\)
\( = 1(2\times0 - 3\times1) - (-1)(5\times0 - 3\times1) + 0(5\times1 - 2\times1)\)
\( = 1(-3) - (-1)(-3)\)
\( = -3 - 3 = -6\)
∴ [ \(\vec{b}\ \vec{c}\ \vec{a}\)\ ] = –6
\([ \vec{b}\ \vec{c}\ \vec{a} ] = \begin{vmatrix} b_x & b_y & b_z \\ c_x & c_y & c_z \\ a_x & a_y & a_z \end{vmatrix}\)
Substituting the values:
\([ \vec{b}\ \vec{c}\ \vec{a} ] = \begin{vmatrix} 1 & -1 & 0\\ 5 & 2 & 3\\ 1 & 1 & 0 \end{vmatrix}\)
\( = 1(2\times0 - 3\times1) - (-1)(5\times0 - 3\times1) + 0(5\times1 - 2\times1)\)
\( = 1(-3) - (-1)(-3)\)
\( = -3 - 3 = -6\)
∴ [ \(\vec{b}\ \vec{c}\ \vec{a}\)\ ] = –6
(2) If \( \vec{\alpha} = \hat{i} - 2\hat{j} + 3\hat{k},\
\vec{\beta} = 2\hat{i} - 3\hat{j} + \hat{k},\
\vec{\gamma} = 3\hat{i} + \hat{j} - 2\hat{k} \), find \( \vec{\alpha} \cdot (\vec{\beta} \times \vec{\gamma}) \).
\( \vec{\beta} \times \vec{\gamma} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & -3 & 1 \\
3 & 1 & -2
\end{vmatrix} =
\hat{i}((-3)(-2) - (1)(1)) -
\hat{j}((2)(-2) - (1)(3)) +
\hat{k}((2)(1) - (-3)(3)) \)
\( = \hat{i}(6 - 1) - \hat{j}(-4 - 3) + \hat{k}(2 + 9) \)
\( = 5\hat{i} + 7\hat{j} + 11\hat{k} \)
Now, \( \vec{\alpha} \cdot (\vec{\beta} \times \vec{\gamma}) = (1)(5) + (-2)(7) + (3)(11) \)
\( = 5 - 14 + 33 = 24 \)
∴ \( \vec{\alpha} \cdot (\vec{\beta} \times \vec{\gamma}) = 24 \)
\( = \hat{i}(6 - 1) - \hat{j}(-4 - 3) + \hat{k}(2 + 9) \)
\( = 5\hat{i} + 7\hat{j} + 11\hat{k} \)
Now, \( \vec{\alpha} \cdot (\vec{\beta} \times \vec{\gamma}) = (1)(5) + (-2)(7) + (3)(11) \)
\( = 5 - 14 + 33 = 24 \)
∴ \( \vec{\alpha} \cdot (\vec{\beta} \times \vec{\gamma}) = 24 \)
(3) If \( \vec{a} = -\hat{i} + 2\hat{j} + \hat{k},\
\vec{b} = 3\hat{i} + \hat{j} + 2\hat{k},\
\vec{c} = 2\hat{i} + \hat{j} + 3\hat{k} \), find \([ \vec{c}\ \vec{a}\ \vec{b} ]\).
\([ \vec{c}\ \vec{a}\ \vec{b} ] =
\begin{vmatrix}
2 & 1 & 3 \\
-1 & 2 & 1 \\
3 & 1 & 2
\end{vmatrix}\)
\( = 2(2\times2 - 1\times1) - 1((-1)\times2 - 1\times3) + 3((-1)\times1 - 2\times3)\)
\( = 2(3) - 1(-2 - 3) + 3(-1 - 6)\)
\( = 6 - (-5) + 3(-7)\)
\( = 6 + 5 - 21 = -10\)
∴ [ \(\vec{c}\ \vec{a}\ \vec{b}\)\ ] = –10
\( = 2(2\times2 - 1\times1) - 1((-1)\times2 - 1\times3) + 3((-1)\times1 - 2\times3)\)
\( = 2(3) - 1(-2 - 3) + 3(-1 - 6)\)
\( = 6 - (-5) + 3(-7)\)
\( = 6 + 5 - 21 = -10\)
∴ [ \(\vec{c}\ \vec{a}\ \vec{b}\)\ ] = –10
(4) If \( \vec{a} = -2\hat{i} - 2\hat{j} + 4\hat{k},\
\vec{b} = -2\hat{i} + 4\hat{j} - 2\hat{k},\
\vec{c} = 4\hat{i} - 2\hat{j} - 2\hat{k} \), find \( \vec{a} \cdot (\vec{b} \times \vec{c}) \).
\( \vec{b} \times \vec{c} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-2 & 4 & -2 \\
4 & -2 & -2
\end{vmatrix}
= \hat{i}((4)(-2)-(-2)(-2)) - \hat{j}((-2)(-2)-(-2)(4)) + \hat{k}((-2)(-2)-(4)(4)) \)
\( = \hat{i}(-8-4) - \hat{j}(4+8) + \hat{k}(4-16) \) \( = -12\hat{i} - 12\hat{j} - 12\hat{k} \)
\( \vec{a} \cdot (\vec{b}\times\vec{c}) = (-2)(-12)+(-2)(-12)+(4)(-12)=24+24-48=0 \)
∴ The vectors are coplanar since the triple product = 0.
\( = \hat{i}(-8-4) - \hat{j}(4+8) + \hat{k}(4-16) \) \( = -12\hat{i} - 12\hat{j} - 12\hat{k} \)
\( \vec{a} \cdot (\vec{b}\times\vec{c}) = (-2)(-12)+(-2)(-12)+(4)(-12)=24+24-48=0 \)
∴ The vectors are coplanar since the triple product = 0.
(5) If the vectors \( x\hat{i} - 4\hat{j} + 5\hat{k},\ \hat{i} + 2\hat{j} + \hat{k},\ 2\hat{i} - \hat{j} + \hat{k} \) are coplanar, find \(x\).
Coplanarity ⇒ scalar triple product = 0.
\( \begin{vmatrix} x & -4 & 5\\ 1 & 2 & 1\\ 2 & -1 & 1 \end{vmatrix} = 0 \)
\( = x(2\times1 - 1\times(-1)) - (-4)(1\times1 - 1\times2) + 5(1\times(-1) - 2\times2)\)
\( = x(2+1) +4(1-2) +5(-1-4)\)
\( = 3x -4 -25 = 0\)
\( ⇒ 3x = 29 ⇒ x = \tfrac{29}{3} \)
∴ x = 29 ⁄ 3
\( \begin{vmatrix} x & -4 & 5\\ 1 & 2 & 1\\ 2 & -1 & 1 \end{vmatrix} = 0 \)
\( = x(2\times1 - 1\times(-1)) - (-4)(1\times1 - 1\times2) + 5(1\times(-1) - 2\times2)\)
\( = x(2+1) +4(1-2) +5(-1-4)\)
\( = 3x -4 -25 = 0\)
\( ⇒ 3x = 29 ⇒ x = \tfrac{29}{3} \)
∴ x = 29 ⁄ 3
(6) If \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k},\
\vec{b} = \hat{i} + 2\hat{j} - 3\hat{k},\
\vec{c} = 3\hat{i} + \lambda\hat{j} + 5\hat{k} \) are coplanar, find \( \lambda \).
\(
\begin{vmatrix}
2 & -1 & 1\\
1 & 2 & -3\\
3 & \lambda & 5
\end{vmatrix} = 0
\)
\( = 2(2\times5 - (-3)\lambda) - (-1)(1\times5 - (-3)\times3) + 1(1\times\lambda - 2\times3) \)
\( = 2(10 + 3\lambda) + (5+9) + (\lambda -6) = 20 + 6\lambda +14 + \lambda -6 = 28 + 7\lambda = 0 \)
\( ⇒ \lambda = -4 \)
∴ λ = –4
\( = 2(2\times5 - (-3)\lambda) - (-1)(1\times5 - (-3)\times3) + 1(1\times\lambda - 2\times3) \)
\( = 2(10 + 3\lambda) + (5+9) + (\lambda -6) = 20 + 6\lambda +14 + \lambda -6 = 28 + 7\lambda = 0 \)
\( ⇒ \lambda = -4 \)
∴ λ = –4
(7) The vectors determining the sides of a parallelepiped are:
(i) \( \hat{i} + \hat{j} + \hat{k},\ \hat{k},\ 3\hat{i} - \hat{j} + 2\hat{k} \)
(ii) \( 2\hat{i} - \hat{j} + \hat{k},\ \hat{i} + 2\hat{j} - 3\hat{k},\ 3\hat{i} - 4\hat{j} + 5\hat{k} \)
(iii) \( \hat{i} + \hat{j} + \hat{k},\ \hat{i} + 2\hat{j} + 2\hat{k},\ \hat{i} - 2\hat{j} + 4\hat{k} \).
Find the volume of each parallelepiped.
(i) \( \hat{i} + \hat{j} + \hat{k},\ \hat{k},\ 3\hat{i} - \hat{j} + 2\hat{k} \)
(ii) \( 2\hat{i} - \hat{j} + \hat{k},\ \hat{i} + 2\hat{j} - 3\hat{k},\ 3\hat{i} - 4\hat{j} + 5\hat{k} \)
(iii) \( \hat{i} + \hat{j} + \hat{k},\ \hat{i} + 2\hat{j} + 2\hat{k},\ \hat{i} - 2\hat{j} + 4\hat{k} \).
Find the volume of each parallelepiped.
Volume = |scalar triple product|.
(i)\( \begin{vmatrix} 1 & 1 & 1\\ 0 & 0 & 1\\ 3 & -1 & 2 \end{vmatrix} = 1(0×2-1×(-1)) - 1(0×2-1×3) + 1(0×(-1)-0×3) = (1) - (-3) + 0 = 4 \)
Vol = 4 cubic units
(ii)\( \begin{vmatrix} 2 & -1 & 1\\ 1 & 2 & -3\\ 3 & -4 & 5 \end{vmatrix} = 2(2×5 - (-3)(-4)) - (-1)(1×5 - (-3)×3) + 1(1×(-4) - 2×3) = 2(10-12)+1(5+9)+(-4-6) = 2(-2)+14-10=0 \)
Vol = 0 → Coplanar vectors
(iii)\( \begin{vmatrix} 1 & 1 & 1\\ 1 & 2 & 2\\ 1 & -2 & 4 \end{vmatrix} = 1(2×4 - 2×(-2)) - 1(1×4 - 2×1) + 1(1×(-2) - 2×1) = 1(8+4) - (4-2) + (-2-2) = 12 - 2 - 4 = 6 \)
Vol = 6 cubic units
(i)\( \begin{vmatrix} 1 & 1 & 1\\ 0 & 0 & 1\\ 3 & -1 & 2 \end{vmatrix} = 1(0×2-1×(-1)) - 1(0×2-1×3) + 1(0×(-1)-0×3) = (1) - (-3) + 0 = 4 \)
Vol = 4 cubic units
(ii)\( \begin{vmatrix} 2 & -1 & 1\\ 1 & 2 & -3\\ 3 & -4 & 5 \end{vmatrix} = 2(2×5 - (-3)(-4)) - (-1)(1×5 - (-3)×3) + 1(1×(-4) - 2×3) = 2(10-12)+1(5+9)+(-4-6) = 2(-2)+14-10=0 \)
Vol = 0 → Coplanar vectors
(iii)\( \begin{vmatrix} 1 & 1 & 1\\ 1 & 2 & 2\\ 1 & -2 & 4 \end{vmatrix} = 1(2×4 - 2×(-2)) - 1(1×4 - 2×1) + 1(1×(-2) - 2×1) = 1(8+4) - (4-2) + (-2-2) = 12 - 2 - 4 = 6 \)
Vol = 6 cubic units
(8) Find the volume of a parallelepiped whose adjacent edges are given by
\( \vec{a} = \hat{i} + \hat{j} + \hat{k},\
\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k},\
\vec{c} = \hat{i} - \hat{j} + \hat{k} \).
Volume = |\( \vec{a} \cdot (\vec{b} \times \vec{c}) \)|
\( \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(3×1 - 4×(-1)) - \hat{j}(2×1 - 4×1) + \hat{k}(2×(-1) - 3×1) \)
\( = \hat{i}(3+4) - \hat{j}(2-4) + \hat{k}(-2-3) = 7\hat{i} + 2\hat{j} - 5\hat{k} \)
\( \vec{a} \cdot (\vec{b}\times\vec{c}) = (1)(7) + (1)(2) + (1)(-5) = 4 \)
∴ Volume = 4 cubic units
\( \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(3×1 - 4×(-1)) - \hat{j}(2×1 - 4×1) + \hat{k}(2×(-1) - 3×1) \)
\( = \hat{i}(3+4) - \hat{j}(2-4) + \hat{k}(-2-3) = 7\hat{i} + 2\hat{j} - 5\hat{k} \)
\( \vec{a} \cdot (\vec{b}\times\vec{c}) = (1)(7) + (1)(2) + (1)(-5) = 4 \)
∴ Volume = 4 cubic units
(9) If the vectors \( \vec{a} = \hat{i} - 2\hat{j} + \hat{k},\
\vec{b} = 2\hat{i} + \hat{j} - 3\hat{k},\
\vec{c} = 3\hat{i} - \hat{j} + 2\hat{k} \) are given, find the volume of the parallelepiped formed by them.
Volume = |\( \vec{a} \cdot (\vec{b} \times \vec{c}) \)|
\( \begin{vmatrix} 1 & -2 & 1\\ 2 & 1 & -3\\ 3 & -1 & 2 \end{vmatrix} = 1(1×2 - (-3)(-1)) - (-2)(2×2 - (-3)×3) + 1(2×(-1) - 1×3) \)
\( = (2 - 3) + 2(4 + 9) + (-2 - 3) = (-1) + 26 - 5 = 20 \)
∴ Volume = 20 cubic units
\( \begin{vmatrix} 1 & -2 & 1\\ 2 & 1 & -3\\ 3 & -1 & 2 \end{vmatrix} = 1(1×2 - (-3)(-1)) - (-2)(2×2 - (-3)×3) + 1(2×(-1) - 1×3) \)
\( = (2 - 3) + 2(4 + 9) + (-2 - 3) = (-1) + 26 - 5 = 20 \)
∴ Volume = 20 cubic units
(10) Find the scalar triple product of vectors
\( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k},\
\vec{b} = 3\hat{i} + \hat{j} - \hat{k},\
\vec{c} = \hat{i} + \hat{j} + \hat{k} \).
\(
\begin{vmatrix}
1 & -2 & 3\\
3 & 1 & -1\\
1 & 1 & 1
\end{vmatrix}
= 1(1×1 - (-1)×1) - (-2)(3×1 - (-1)×1) + 3(3×1 - 1×1)
\)
\( = (1+1) + 2(3+1) + 3(3-1) = 2 + 8 + 6 = 16 \)
∴ Scalar triple product = 16
\( = (1+1) + 2(3+1) + 3(3-1) = 2 + 8 + 6 = 16 \)
∴ Scalar triple product = 16
(11) If \( \vec{a} = \hat{i} + \hat{j} + \hat{k},\
\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k},\
\vec{c} = 3\hat{i} + 2\hat{j} + \lambda\hat{k} \) are coplanar, find \( \lambda \).
Coplanar ⇒ scalar triple product = 0.
\( \begin{vmatrix} 1 & 1 & 1\\ 2 & 1 & 3\\ 3 & 2 & \lambda \end{vmatrix} = 0 \)
\( = 1(1×\lambda - 3×2) - 1(2×\lambda - 3×3) + 1(2×2 - 1×3) = (\lambda - 6) - (2\lambda - 9) + (4 - 3) \)
\( = \lambda - 6 - 2\lambda + 9 + 1 = -\lambda + 4 = 0 \)
\( ⇒ \lambda = 4 \)
∴ λ = 4
\( \begin{vmatrix} 1 & 1 & 1\\ 2 & 1 & 3\\ 3 & 2 & \lambda \end{vmatrix} = 0 \)
\( = 1(1×\lambda - 3×2) - 1(2×\lambda - 3×3) + 1(2×2 - 1×3) = (\lambda - 6) - (2\lambda - 9) + (4 - 3) \)
\( = \lambda - 6 - 2\lambda + 9 + 1 = -\lambda + 4 = 0 \)
\( ⇒ \lambda = 4 \)
∴ λ = 4
(12) If the vectors \( \vec{a} = a\hat{i} + a\hat{j} + c\hat{k},\
\vec{b} = \hat{i} + \hat{k},\
\vec{c} = c\hat{i} + c\hat{j} + b\hat{k} \) are coplanar, show that \( c^2 = ab \).
Coplanar ⇒ \( [\vec{a}\ \vec{b}\ \vec{c}] = 0 \)
\( \begin{vmatrix} a & a & c\\ 1 & 0 & 1\\ c & c & b \end{vmatrix} = 0 \)
\( = a(0×b - 1×c) - a(1×b - 1×c) + c(1×c - 0×c) \)
\( = a(-c) - a(b - c) + c^2 = -ac - ab + ac + c^2 = c^2 - ab = 0 \)
∴ \( c^2 = ab \)
\( \begin{vmatrix} a & a & c\\ 1 & 0 & 1\\ c & c & b \end{vmatrix} = 0 \)
\( = a(0×b - 1×c) - a(1×b - 1×c) + c(1×c - 0×c) \)
\( = a(-c) - a(b - c) + c^2 = -ac - ab + ac + c^2 = c^2 - ab = 0 \)
∴ \( c^2 = ab \)
(13) Let \( \vec{a} = \hat{i} + \hat{j} + \hat{k},\
\vec{b} = \hat{i},\
\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k} \).
If \( c_1 = 1, c_2 = 2 \), find \( c_3 \) so that \( \vec{a}, \vec{b}, \vec{c} \) are coplanar.
For coplanarity:
\(
\begin{vmatrix}
1 & 1 & 1\\
1 & 0 & 0\\
1 & 2 & c_3
\end{vmatrix} = 0
\)
\( = 1(0×c_3 - 0×2) - 1(1×c_3 - 0×1) + 1(1×2 - 0×1) = 0 - c_3 + 2 = 0 \)
\( ⇒ c_3 = 2 \)
∴ \( c_3 = 2 \)
\( = 1(0×c_3 - 0×2) - 1(1×c_3 - 0×1) + 1(1×2 - 0×1) = 0 - c_3 + 2 = 0 \)
\( ⇒ c_3 = 2 \)
∴ \( c_3 = 2 \)
(14) Find the value of \( x \) such that the points
\( A(3, 2, 1), B(4, x, 5), C(4, 2, -2), D(6, 5, -1) \) are coplanar.
Four points are coplanar ⇒
\( [\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}] = 0 \)
\( \overrightarrow{AB} = (4-3)\hat{i} + (x-2)\hat{j} + (5-1)\hat{k} = \hat{i} + (x-2)\hat{j} + 4\hat{k} \)
\( \overrightarrow{AC} = (4-3)\hat{i} + (2-2)\hat{j} + (-2-1)\hat{k} = \hat{i} - 3\hat{k} \)
\( \overrightarrow{AD} = (6-3)\hat{i} + (5-2)\hat{j} + (-1-1)\hat{k} = 3\hat{i} + 3\hat{j} - 2\hat{k} \)
\( \begin{vmatrix} 1 & (x-2) & 4\\ 1 & 0 & -3\\ 3 & 3 & -2 \end{vmatrix} = 0 \)
Expanding: \( = 1(0×(-2) - (-3)×3) - (x-2)(1×(-2) - (-3)×3) + 4(1×3 - 0×3) = 9 - (x-2)(7) + 12 = 21 - 7x + 14 = 0 \)
\( ⇒ -7x + 35 = 0 ⇒ x = 5 \)
∴ \( x = 5 \)
\( \overrightarrow{AB} = (4-3)\hat{i} + (x-2)\hat{j} + (5-1)\hat{k} = \hat{i} + (x-2)\hat{j} + 4\hat{k} \)
\( \overrightarrow{AC} = (4-3)\hat{i} + (2-2)\hat{j} + (-2-1)\hat{k} = \hat{i} - 3\hat{k} \)
\( \overrightarrow{AD} = (6-3)\hat{i} + (5-2)\hat{j} + (-1-1)\hat{k} = 3\hat{i} + 3\hat{j} - 2\hat{k} \)
\( \begin{vmatrix} 1 & (x-2) & 4\\ 1 & 0 & -3\\ 3 & 3 & -2 \end{vmatrix} = 0 \)
Expanding: \( = 1(0×(-2) - (-3)×3) - (x-2)(1×(-2) - (-3)×3) + 4(1×3 - 0×3) = 9 - (x-2)(7) + 12 = 21 - 7x + 14 = 0 \)
\( ⇒ -7x + 35 = 0 ⇒ x = 5 \)
∴ \( x = 5 \)
