CBSE Class 12 Mathematics Vector Algebra Assignment Set 03

Chapter - 10

Vector Algebra

I. Scalar Triple Product

Consider three vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \). The dot product of \( \vec{a} \times \vec{b} \) with \( \vec{c} \), written as \( (\vec{a} \times \vec{b}) \cdot \vec{c} \), represents the scalar triple product of these vectors. It is compactly denoted using box notation as \( [\vec{a}\;\vec{b}\;\vec{c}] \).

\[ [\vec{a}\;\vec{b}\;\vec{c}] = (\vec{a} \times \vec{b}) \cdot \vec{c} \]

II. Geometrical Interpretation of a Scalar Triple Product

Let three concurrent coterminous edges \( OA \), \( OB \), and \( OC \) of a parallelepiped be represented by vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) respectively. The cross product \( \vec{b} \times \vec{c} \) represents the vector area of the base face formed by \( \vec{b} \) and \( \vec{c} \). The altitude of the parallelepiped corresponds to the scalar projection of edge vector \( \vec{a} \) onto the normal vector \( \vec{b} \times \vec{c} \).

Magnitude of this projection is given by:

\[ \text{Height} = \frac{|\vec{a} \cdot (\vec{b} \times \vec{c})|}{|\vec{b} \times \vec{c}|} \]

The volume of the parallelepiped is calculated as:

\[ \text{Volume} = (\text{Area of base}) \times (\text{Height}) = |\vec{b} \times \vec{c}| \left[ \frac{|\vec{a} \cdot (\vec{b} \times \vec{c})|}{|\vec{b} \times \vec{c}|} \right] = |\vec{a} \cdot (\vec{b} \times \vec{c})| \]

{ Modulus has been taken as area is always positive }

Thus, if \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) represent the three co-terminus edges of a parallelepiped, then its volume is \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) or \( [\vec{a}\;\vec{b}\;\vec{c}] \).

III. Scalar Triple Product in Terms of Rectangular Components

Let \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \), \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), and \( \vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k} \).

Then:

\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = (b_2 c_3 - b_3 c_2)\hat{i} - (b_1 c_3 - b_3 c_1)\hat{j} + (b_1 c_2 - b_2 c_1)\hat{k} \]

Taking the scalar product with \( \vec{a} \):

\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = a_1(b_2 c_3 - b_3 c_2) - a_2(b_1 c_3 - b_3 c_1) + a_3(b_1 c_2 - b_2 c_1) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \]

Remarks:

• If for any three vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \), \( [\vec{a}\;\vec{b}\;\vec{c}] = 0 \), then the volume of the parallelepiped formed by these coterminous edges is zero. This situation is possible if and only if \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are coplanar vectors.
• Thus, \( [\vec{a}\;\vec{b}\;\vec{c}] = 0 \iff \vec{a}, \vec{b}, \vec{c} \) are co-planar.

IV. Properties of Scalar Triple Product

1. Cyclic Permutations: If \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are any three vectors, then:

\[ [\vec{a}\;\vec{b}\;\vec{c}] = [\vec{b}\;\vec{c}\;\vec{a}] = [\vec{c}\;\vec{a}\;\vec{b}] \]

Proof: Interchanging two rows of a determinant multiplies its value by \( -1 \). Performing two consecutive row interchanges leaves the determinant value unchanged:

\[ [\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = - \begin{vmatrix} b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = (-1)^2 \begin{vmatrix} b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ a_1 & a_2 & a_3 \end{vmatrix} = [\vec{b}\;\vec{c}\;\vec{a}] \quad \text{---(i)} \]

Similarly, it can be verified that \( [\vec{a}\;\vec{b}\;\vec{c}] = [\vec{c}\;\vec{a}\;\vec{b}] \) ---(ii).

From (i) and (ii), we obtain:

\[ [\vec{a}\;\vec{b}\;\vec{c}] = [\vec{b}\;\vec{c}\;\vec{a}] = [\vec{c}\;\vec{a}\;\vec{b}] \]

\implies If \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are cyclically permuted, the value of the scalar triple product remains unaltered.

2. Interchanging Dot and Cross: In a scalar triple product, the positions of the dot and cross can be interchanged, provided the cyclic order of vectors remains the same.

Proof: Since \( [\vec{a}\;\vec{b}\;\vec{c}] = [\vec{c}\;\vec{a}\;\vec{b}] \)

\implies \( \vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{c} \cdot (\vec{a} \times \vec{b}) \)

or \( \vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{c} \cdot (\vec{a} \times \vec{b}) = (\vec{a} \times \vec{b}) \cdot \vec{c} \).

3. Change of Cyclic Order: The value of the scalar triple product remains the same in magnitude, but changes the sign, if the cyclic order of \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) is changed.

Proof: \( [\vec{a}\;\vec{b}\;\vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = -\vec{a} \cdot (\vec{c} \times \vec{b}) = -[\vec{a}\;\vec{c}\;\vec{b}] \).

4. Two Equal Vectors: The scalar triple product of three vectors is zero if any two of the given vectors are equal.

Proof: Let \( \vec{a} = \vec{b} \).

\[ [\vec{a}\;\vec{b}\;\vec{c}] = [\vec{a}\;\vec{a}\;\vec{c}] = (\vec{a} \times \vec{a}) \cdot \vec{c} = \vec{0} \cdot \vec{c} = 0 \]

Similarly, if \( \vec{b} = \vec{c} \) or \( \vec{c} = \vec{a} \), then \( [\vec{a}\;\vec{b}\;\vec{c}] = 0 \).

5. Scalar Multiplication: For any three vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) and scalar \( \lambda \), we have:

\[ [\lambda\vec{a}\;\;\vec{b}\;\;\vec{c}] = \lambda [\vec{a}\;\vec{b}\;\vec{c}] \]

Proof: \( [\lambda\vec{a}\;\;\vec{b}\;\;\vec{c}] = (\lambda\vec{a} \times \vec{b}) \cdot \vec{c} = \lambda(\vec{a} \times \vec{b}) \cdot \vec{c} = \lambda [\vec{a}\;\vec{b}\;\vec{c}] \).

6. Parallel or Collinear Vectors: The scalar triple product of three vectors is zero if any two of them are parallel or collinear.

Proof: Let \( \vec{a} \) be parallel (or collinear) to \( \vec{b} \). Then \( \vec{a} = \lambda\vec{b} \) for some scalar \( \lambda \).

\[ [\vec{a}\;\vec{b}\;\vec{c}] = [\lambda\vec{b}\;\;\vec{b}\;\;\vec{c}] = \lambda [\vec{b}\;\vec{b}\;\vec{c}] = \lambda \cdot 0 = 0 \]

Examples

Question (Example 1). If \( \vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} \), \( \vec{b} = \hat{i} + 2\hat{j} - 3\hat{k} \) and \( \vec{c} = 3\hat{i} + 4\hat{j} - \hat{k} \) then find \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) and \( (\vec{a} \times \vec{b}) \cdot \vec{c} \). Is \( \vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c} \)?
Answer: First compute \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) by expanding the determinant formed from the coordinates of the three vectors:
\( \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -3 \\ 3 & 4 & -1 \end{vmatrix} \)
Expanding across row 1:
\( = 2[2(-1) - (-3)(4)] - (-3)[1(-1) - (-3)(3)] + 4[1(4) - 2(3)] \)
\( = 2(-2 + 12) + 3(-1 + 9) + 4(4 - 6) \)
\( = 2(10) + 3(8) + 4(-2) = 20 + 24 - 8 = 36 \).
Next calculate \( (\vec{a} \times \vec{b}) \cdot \vec{c} = \vec{c} \cdot (\vec{a} \times \vec{b}) \):
\( (\vec{a} \times \vec{b}) \cdot \vec{c} = \begin{vmatrix} 3 & 4 & -1 \\ 2 & -3 & 4 \\ 1 & 2 & -3 \end{vmatrix} \)
Expanding across row 1:
\( = 3[(-3)(-3) - 4(2)] - 4[2(-3) - 4(1)] - 1[2(2) - (-3)(1)] \)
\( = 3(9 - 8) - 4(-6 - 4) - 1(4 + 3) \)
\( = 3(1) - 4(-10) - 1(7) = 3 + 40 - 7 = 36 \).
Because both computations arrive at 36, we conclude that \( \vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c} \).
In simple words: Evaluating the determinant in both orders gives 36, proving both calculations yield the exact same scalar value.

Exam Tip: This illustrates the fundamental property that dot and cross operators can interchange places without altering the scalar triple product.

Question (Example 2). Find the volume of the parallelopipped whose edges are \( \vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} \), \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \) and \( \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k} \)
Answer: The volume of a parallelepiped determined by three concurrent edges corresponds to the magnitude of their scalar triple product \( |[\vec{a}\;\vec{b}\;\vec{c}]| \):
\( [\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 2 & -1 & 2 \end{vmatrix} \)
Expanding along the top row:
\( = 2[2(2) - (-1)(-1)] - (-3)[1(2) - (-1)(2)] + 4[1(-1) - 2(2)] \)
\( = 2(4 - 1) + 3(2 + 2) + 4(-1 - 4) \)
\( = 2(3) + 3(4) + 4(-5) = 6 + 12 - 20 = -2 \).
Because geometric volume is strictly non-negative, taking the absolute value gives:
\( \text{Volume} = |-2| = 2 \text{ cubic units} \).
In simple words: Work out the 3 by 3 determinant and take the positive magnitude to find the required volume.

Exam Tip: Never write a negative answer for volume; clearly state that you are taking the absolute value.

Question (Example 3). Show that the vectors \( \vec{a} = -4\hat{i} - 6\hat{j} - 2\hat{k} \), \( \vec{b} = -\hat{i} + 4\hat{j} + 3\hat{k} \) and \( \vec{c} = -8\hat{i} - \hat{j} + 3\hat{k} \) are co-planar
Answer: Three vectors are coplanar if and only if their scalar triple product is equal to zero:
\( [\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} -4 & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} \)
Expanding across row 1:
\( = -4[4(3) - 3(-1)] - (-6)[(-1)(3) - 3(-8)] - 2[(-1)(-1) - 4(-8)] \)
\( = -4(12 + 3) + 6(-3 + 24) - 2(1 + 32) \)
\( = -4(15) + 6(21) - 2(33) \)
\( = -60 + 126 - 66 = 0 \).
Since \( [\vec{a}\;\vec{b}\;\vec{c}] = 0 \), the vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are coplanar.
In simple words: The component determinant evaluates to zero, proving the vectors lie in the same plane.

Exam Tip: Always state the coplanarity condition \( [\vec{a}\;\vec{b}\;\vec{c}] = 0 \) before performing determinant expansion.

Question (Example 4). Show that the vectors \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \), \( \vec{b} = -2\hat{i} + 3\hat{j} - 4\hat{k} \) and \( \vec{c} = \hat{i} - 3\hat{j} + \lambda\hat{k} \) are co-planar if λ = 5
Answer: Vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) lie in one plane when their scalar triple product is zero:
\( [\vec{a}\;\vec{b}\;\vec{c}] = 0 \)

\implies \( \begin{vmatrix} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & \lambda \end{vmatrix} = 0 \).
Expand the determinant across row 1:
\( 1[3(\lambda) - (-4)(-3)] - (-2)[-2(\lambda) - (-4)(1)] + 3[(-2)(-3) - 3(1)] = 0 \)

\implies \( (3\lambda - 12) + 2(-2\lambda + 4) + 3(6 - 3) = 0 \)

\implies \( 3\lambda - 12 - 4\lambda + 8 + 9 = 0 \)

\implies \( -\lambda + 5 = 0 \)

\implies \( \lambda = 5 \).
Thus, the vectors are coplanar if \( \lambda = 5 \).
In simple words: Setting the determinant to zero gives a linear equation that simplifies to lambda = 5.

Exam Tip: Be mindful of signs when expanding minors containing negative scalar components.

Question (Example 5). Show that four points with position vectors \( 6\hat{i} - 7\hat{j} \), \( 16\hat{i} - 19\hat{j} - 4\hat{k} \), \( 3\hat{j} - 6\hat{k} \) and \( 2\hat{i} + 5\hat{j} + 10\hat{k} \) are not co-planar
Answer: Designate the four given points as \( A \), \( B \), \( C \), and \( D \) with respective position vectors:
\( \vec{OA} = 6\hat{i} - 7\hat{j} \)
\( \vec{OB} = 16\hat{i} - 19\hat{j} - 4\hat{k} \)
\( \vec{OC} = 3\hat{j} - 6\hat{k} \)
\( \vec{OD} = 2\hat{i} + 5\hat{j} + 10\hat{k} \).
Construct three displacement vectors sharing point \( A \) as common initial vertex:
\( \vec{a} = \vec{AB} = \vec{OB} - \vec{OA} = (16 - 6)\hat{i} + (-19 - (-7))\hat{j} + (-4 - 0)\hat{k} = 10\hat{i} - 12\hat{j} - 4\hat{k} \)
\( \vec{b} = \vec{AC} = \vec{OC} - \vec{OA} = (0 - 6)\hat{i} + (3 - (-7))\hat{j} + (-6 - 0)\hat{k} = -6\hat{i} + 10\hat{j} - 6\hat{k} \)
\( \vec{c} = \vec{AD} = \vec{OD} - \vec{OA} = (2 - 6)\hat{i} + (5 - (-7))\hat{j} + (10 - 0)\hat{k} = -4\hat{i} + 12\hat{j} + 10\hat{k} \).
The four points are coplanar if and only if \( [\vec{a}\;\vec{b}\;\vec{c}] = 0 \).
Now evaluate the determinant:
\( [\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} 10 & -12 & -4 \\ -6 & 10 & -6 \\ -4 & 12 & 10 \end{vmatrix} \)
Expanding along row 1:
\( = 10[10(10) - (-6)(12)] - (-12)[(-6)(10) - (-6)(-4)] - 4[(-6)(12) - 10(-4)] \)
\( = 10(100 + 72) + 12(-60 - 24) - 4(-72 + 40) \)
\( = 10(172) + 12(-84) - 4(-32) \)
\( = 1720 - 1008 + 128 = 840 \neq 0 \).
Because \( [\vec{a}\;\vec{b}\;\vec{c}] \neq 0 \), the three vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \) are non-coplanar.
Hence, points \( A, B, C, \text{and } D \) are not coplanar.
In simple words: Form three direction vectors by subtracting point A from the other points; since their box product is 840 and not zero, they do not share a common plane.

Exam Tip: Remember to supply \( 0\hat{k} \) for point A and \( 0\hat{i} \) for point C when writing out their position coordinates.

Question (Example 6). For any three vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) , prove that \( [\vec{a}+\vec{b}\;\;\vec{b}+\vec{c}\;\;\vec{c}+\vec{a}] = 2 [\vec{a}\;\vec{b}\;\vec{c}] \)
Answer: By definition of scalar triple product:
\( [\vec{a} + \vec{b}\;\;\vec{b} + \vec{c}\;\;\vec{c} + \vec{a}] = [(\vec{a} + \vec{b}) \times (\vec{b} + \vec{c})] \cdot (\vec{c} + \vec{a}) \).
First expand the cross product across the sum:
\( (\vec{a} + \vec{b}) \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{b} + \vec{b} \times \vec{c} \).
Since \( \vec{b} \times \vec{b} = \vec{0} \), this reduces to:
\( = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c} \).
Now take the scalar product with \( (\vec{c} + \vec{a}) \):
\( = (\vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c}) \cdot (\vec{c} + \vec{a}) \)
\( = (\vec{a} \times \vec{b}) \cdot \vec{c} + (\vec{a} \times \vec{b}) \cdot \vec{a} + (\vec{a} \times \vec{c}) \cdot \vec{c} + (\vec{a} \times \vec{c}) \cdot \vec{a} + (\vec{b} \times \vec{c}) \cdot \vec{c} + (\vec{b} \times \vec{c}) \cdot \vec{a} \).
A scalar triple product with any two equal vectors is identically zero:
\( (\vec{a} \times \vec{b}) \cdot \vec{a} = 0 \), \( (\vec{a} \times \vec{c}) \cdot \vec{c} = 0 \), \( (\vec{a} \times \vec{c}) \cdot \vec{a} = 0 \), and \( (\vec{b} \times \vec{c}) \cdot \vec{c} = 0 \).
The remaining expression simplifies to:
\( = (\vec{a} \times \vec{b}) \cdot \vec{c} + (\vec{b} \times \vec{c}) \cdot \vec{a} \)
\( = [\vec{a}\;\vec{b}\;\vec{c}] + [\vec{b}\;\vec{c}\;\vec{a}] \).
Using the cyclic invariance property \( [\vec{b}\;\vec{c}\;\vec{a}] = [\vec{a}\;\vec{b}\;\vec{c}] \):
\( = [\vec{a}\;\vec{b}\;\vec{c}] + [\vec{a}\;\vec{b}\;\vec{c}] = 2 [\vec{a}\;\vec{b}\;\vec{c}] \).
Hence proved.
In simple words: Multiplying out the brackets leaves only two non-zero cyclic terms, and adding them together gives twice the original triple product.

Exam Tip: Explicitly mention that terms containing two identical vectors vanish because their cross product is perpendicular to either vector.

Exercise 1

Question 1. If \( \vec{a} = 7\hat{i} - 2\hat{j} + 3\hat{k} \), \( \vec{b} = \hat{i} - \hat{j} + 2\hat{k} \) and \( \vec{c} = 3\hat{i} + 8\hat{j} \), then find \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) and \( (\vec{a} \times \vec{b}) \cdot \vec{c} \) Also Find whether \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) and \( (\vec{a} \times \vec{b}) \cdot \vec{c} \) are equal
Answer: First compute \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) using a \( 3 \times 3 \) determinant where \( \vec{c} = 3\hat{i} + 8\hat{j} + 0\hat{k} \):
\( \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} 7 & -2 & 3 \\ 1 & -1 & 2 \\ 3 & 8 & 0 \end{vmatrix} \)
Expanding along row 3:
\( = 3[(-2)(2) - 3(-1)] - 8[7(2) - 3(1)] + 0 \)
\( = 3(-4 + 3) - 8(14 - 3) \)
\( = 3(-1) - 8(11) = -3 - 88 = -91 \).
Next compute \( (\vec{a} \times \vec{b}) \cdot \vec{c} = \vec{c} \cdot (\vec{a} \times \vec{b}) \):
\( (\vec{a} \times \vec{b}) \cdot \vec{c} = \begin{vmatrix} 3 & 8 & 0 \\ 7 & -2 & 3 \\ 1 & -1 & 2 \end{vmatrix} \)
Expanding along row 1:
\( = 3[(-2)(2) - 3(-1)] - 8[7(2) - 3(1)] + 0 \)
\( = 3(-4 + 3) - 8(14 - 3) = 3(-1) - 8(11) = -3 - 88 = -91 \).
Both calculations yield \( -91 \).
Therefore, \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) and \( (\vec{a} \times \vec{b}) \cdot \vec{c} \) are equal.
In simple words: Expanding both determinant setups gives -91, confirming the two expressions have the exact same value.

Exam Tip: Expanding along the row containing zero (row 3 for the first determinant, row 1 for the second) minimizes arithmetic steps.

Question 2. If \( \vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} \), \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \) and \( \vec{c} = 3\hat{i} - \hat{j} + 2\hat{k} \), then
(i) find \( [\vec{a}\;\vec{b}\;\vec{c}] \)
(ii) find \( [\vec{a}+\vec{b}\;\;\vec{b}+\vec{c}\;\;\vec{c}+\vec{a}] \)
Answer:
(i) Evaluate the scalar triple product \( [\vec{a}\;\vec{b}\;\vec{c}] \) by forming a determinant of the vector components:
\( [\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} \)
Expanding across row 1:
\( = 2[2(2) - (-1)(-1)] - (-3)[1(2) - (-1)(3)] + 4[1(-1) - 2(3)] \)
\( = 2(4 - 1) + 3(2 + 3) + 4(-1 - 6) \)
\( = 2(3) + 3(5) + 4(-7) = 6 + 15 - 28 = -7 \).

(ii) Using the known vector identity \( [\vec{a} + \vec{b}\;\;\vec{b} + \vec{c}\;\;\vec{c} + \vec{a}] = 2 [\vec{a}\;\vec{b}\;\vec{c}] \):
Substitute the result from part (i):
\( [\vec{a} + \vec{b}\;\;\vec{b} + \vec{c}\;\;\vec{c} + \vec{a}] = 2(-7) = -14 \).
In simple words: The first box product computes to -7, and using the double identity gives -14 for the second part.

Exam Tip: Quoting \( [\vec{a}+\vec{b}\;\vec{b}+\vec{c}\;\vec{c}+\vec{a}] = 2[\vec{a}\;\vec{b}\;\vec{c}] \) directly receives full credit without requiring full expansion.

Question 3. Find the volumes of the following parallelopipeds whose three co-terminus edges are:
(i) \( \vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} \), \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \) and \( \vec{c} = \hat{i} + 2\hat{j} - \hat{k} \)
(ii) \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \), \( \vec{b} = 2\hat{i} + \hat{j} - \hat{k} \), and \( \vec{c} = 2\hat{i} + \hat{j} - \hat{k} \)
Answer:
(i) The volume of the parallelepiped is given by the absolute value of the scalar triple product \( |[\vec{a}\;\vec{b}\;\vec{c}]| \):
\( [\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} 2 & -3 & 4 \\ 3 & -1 & 2 \\ 1 & 2 & -1 \end{vmatrix} \)
Expanding along the first row:
\( = 2[(-1)(-1) - 2(2)] - (-3)[3(-1) - 2(1)] + 4[3(2) - (-1)(1)] \)
\( = 2(1 - 4) + 3(-3 - 2) + 4(6 + 1) \)
\( = 2(-3) + 3(-5) + 4(7) = -6 - 15 + 28 = 7 \).
Thus, the volume is \( |7| = 7 \text{ cubic units} \).

(ii) For the given edges, vectors \( \vec{b} \) and \( \vec{c} \) are identical: \( \vec{b} = \vec{c} = 2\hat{i} + \hat{j} - \hat{k} \).
Whenever two coterminous edge vectors coincide, the scalar triple product is zero:
\( [\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} 1 & -2 & 3 \\ 2 & 1 & -1 \\ 2 & 1 & -1 \end{vmatrix} = 0 \).
Hence, the volume of the parallelepiped is \( 0 \text{ cubic units} \).
In simple words: Part (i) evaluates to 7 cubic units, while part (ii) gives 0 cubic units because two edges are identical.

Exam Tip: If two vectors are identical, cite the property that identical rows make the determinant vanish rather than doing long calculations.

Question 4. Show that
(i) the vectors \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} + 2\hat{j} - 3\hat{k} \), and \( \vec{c} = 3\hat{i} - 4\hat{j} + 5\hat{k} \) are coplanar
(ii) the vectors \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \), \( \vec{b} = -2\hat{i} + 3\hat{j} - 4\hat{k} \), and \( \vec{c} = \hat{i} - 3\hat{j} + 5\hat{k} \) are coplanar
Answer:
(i) Three vectors are coplanar when their scalar triple product is zero:
\( [\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & -4 & 5 \end{vmatrix} \)
Expanding across row 1:
\( = 2[2(5) - (-3)(-4)] - (-1)[1(5) - (-3)(3)] + 1[1(-4) - 2(3)] \)
\( = 2(10 - 12) + 1(5 + 9) + 1(-4 - 6) \)
\( = 2(-2) + 1(14) + 1(-10) = -4 + 14 - 10 = 0 \).
Since \( [\vec{a}\;\vec{b}\;\vec{c}] = 0 \), the vectors are coplanar.

(ii) Compute the scalar triple product for the second set of vectors:
\( [\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix} \)
Expanding across row 1:
\( = 1[3(5) - (-4)(-3)] - (-2)[(-2)(5) - (-4)(1)] + 3[(-2)(-3) - 3(1)] \)
\( = 1(15 - 12) + 2(-10 + 4) + 3(6 - 3) \)
\( = 1(3) + 2(-6) + 3(3) = 3 - 12 + 9 = 0 \).
Since \( [\vec{a}\;\vec{b}\;\vec{c}] = 0 \), these vectors are also coplanar.
Hence proved.
In simple words: The component determinant equals zero in both parts, which shows that each group of vectors lies in a flat plane.

Exam Tip: Always conclude your solution by writing "Since \( [\vec{a}\;\vec{b}\;\vec{c}] = 0 \), the vectors are coplanar."

Question 5. Find the value of λ if the following vectors are co-planar
(i) \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} + 2\hat{j} - 3\hat{k} \), and \( \vec{c} = 3\hat{i} - \lambda\hat{j} + 5\hat{k} \)
(ii) \( \vec{a} = 2\hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} - \lambda\hat{j} + \hat{k} \), and \( \vec{c} = 5\hat{i} + \hat{j} - 3\hat{k} \)
Answer:
(i) For the vectors to be coplanar, their scalar triple product must equal zero:
\( \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & -\lambda & 5 \end{vmatrix} = 0 \)
Expanding along the first row:
\( 2[2(5) - (-3)(-\lambda)] - (-1)[1(5) - (-3)(3)] + 1[1(-\lambda) - 2(3)] = 0 \)

\implies \( 2(10 - 3\lambda) + 1(5 + 9) + 1(-\lambda - 6) = 0 \)

\implies \( 20 - 6\lambda + 14 - \lambda - 6 = 0 \)

\implies \( 28 - 7\lambda = 0 \)

\implies \( 7\lambda = 28 
\implies \lambda = 4 \).

(ii) Set the determinant of coordinates to zero:
\( \begin{vmatrix} 2 & 1 & 1 \\ 2 & -\lambda & 1 \\ 5 & 1 & -3 \end{vmatrix} = 0 \)
Expanding across row 1:
\( 2[(-\lambda)(-3) - 1(1)] - 1[2(-3) - 1(5)] + 1[2(1) - (-\lambda)(5)] = 0 \)

\implies \( 2(3\lambda - 1) - 1(-6 - 5) + 1(2 + 5\lambda) = 0 \)

\implies \( 6\lambda - 2 + 11 + 2 + 5\lambda = 0 \)

\implies \( 11\lambda + 11 = 0 \)

\implies \( 11\lambda = -11 
\implies \lambda = -1 \).
In simple words: Setting the determinant to zero gives a basic linear equation in lambda, giving lambda = 4 for part (i) and lambda = -1 for part (ii).

Exam Tip: Be careful when simplifying negative signs inside parentheses, especially in \( (-\lambda)(-3) = 3\lambda \).

Question 6. If \( \hat{i} \), \( \hat{j} \) and \( \hat{k} \) are three mutually perpendicular vectors, prove that \( \hat{i} \cdot (\hat{k} \times \hat{j}) = \hat{j} \cdot (\hat{i} \times \hat{k}) = \hat{k} \cdot (\hat{j} \times \hat{i}) = -1 \)
Answer: For a standard right-handed orthogonal unit vector system, cross products follow cyclic order:
\( \hat{j} \times \hat{k} = \hat{i} \implies \hat{k} \times \hat{j} = -\hat{i} \)
\( \hat{k} \times \hat{i} = \hat{j} \implies \hat{i} \times \hat{k} = -\hat{j} \)
\( \hat{i} \times \hat{j} = \hat{k} \implies \hat{j} \times \hat{i} = -\hat{k} \).
Evaluate the three given expressions:
1. \( \hat{i} \cdot (\hat{k} \times \hat{j}) = \hat{i} \cdot (-\hat{i}) = -(\hat{i} \cdot \hat{i}) = -1 \).
2. \( \hat{j} \cdot (\hat{i} \times \hat{k}) = \hat{j} \cdot (-\hat{j}) = -(\hat{j} \cdot \hat{j}) = -1 \).
3. \( \hat{k} \cdot (\hat{j} \times \hat{i}) = \hat{k} \cdot (-\hat{k}) = -(\hat{k} \cdot \hat{k}) = -1 \).
All three terms produce the value \( -1 \).
Hence, \( \hat{i} \cdot (\hat{k} \times \hat{j}) = \hat{j} \cdot (\hat{i} \times \hat{k}) = \hat{k} \cdot (\hat{j} \times \hat{i}) = -1 \).
Hence proved.
In simple words: Because each cross product runs against the normal cyclic direction, it produces a negative unit vector, which dots with itself to make -1.

Exam Tip: State the cyclic rule of unit vectors clearly before substituting the cross products.

Question 7. Show that the four points A, B, C and D with position vectors \( 4\hat{i} + 5\hat{j} + \hat{k} \), \( -(\hat{j} + \hat{k}) \), \( 3\hat{i} + 9\hat{j} + 4\hat{k} \) and \( -4\hat{i} + 4\hat{j} + 4\hat{k} \), respectively are co-planar.
Answer: Let the position vectors of the given points be:
\( \vec{OA} = 4\hat{i} + 5\hat{j} + \hat{k} \)
\( \vec{OB} = 0\hat{i} - \hat{j} - \hat{k} \)
\( \vec{OC} = 3\hat{i} + 9\hat{j} + 4\hat{k} \)
\( \vec{OD} = -4\hat{i} + 4\hat{j} + 4\hat{k} \).
Form three vectors sharing point \( A \) as common initial point:
\( \vec{AB} = \vec{OB} - \vec{OA} = (0 - 4)\hat{i} + (-1 - 5)\hat{j} + (-1 - 1)\hat{k} = -4\hat{i} - 6\hat{j} - 2\hat{k} \)
\( \vec{AC} = \vec{OC} - \vec{OA} = (3 - 4)\hat{i} + (9 - 5)\hat{j} + (4 - 1)\hat{k} = -\hat{i} + 4\hat{j} + 3\hat{k} \)
\( \vec{AD} = \vec{OD} - \vec{OA} = (-4 - 4)\hat{i} + (4 - 5)\hat{j} + (4 - 1)\hat{k} = -8\hat{i} - \hat{j} + 3\hat{k} \).
The four points are coplanar if \( [\vec{AB}\;\vec{AC}\;\vec{AD}] = 0 \):
\( [\vec{AB}\;\vec{AC}\;\vec{AD}] = \begin{vmatrix} -4 & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} \)
Expanding along row 1:
\( = -4[4(3) - 3(-1)] - (-6)[(-1)(3) - 3(-8)] - 2[(-1)(-1) - 4(-8)] \)
\( = -4(12 + 3) + 6(-3 + 24) - 2(1 + 32) \)
\( = -4(15) + 6(21) - 2(33) \)
\( = -60 + 126 - 66 = 0 \).
Because the scalar triple product is zero, the vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \) are coplanar, confirming that points \( A, B, C, \text{and } D \) lie in the same plane.
Hence proved.
In simple words: Find vectors AB, AC, and AD from point A; since their box product equals zero, all four points sit in the same plane.

Exam Tip: Be sure to write \( -(\hat{j} + \hat{k}) \) as \( 0\hat{i} - 1\hat{j} - 1\hat{k} \) to avoid missing the zero coordinate.

Question 8. Find the value of λ if the points A (-1, 4, -3), B = (3, λ, -5), C (-3, 8 -5) and D (-3, 2, 1) are coplanar
Answer: Let the coordinates of the four points be \( A(-1, 4, -3) \), \( B(3, \lambda, -5) \), \( C(-3, 8, -5) \), and \( D(-3, 2, 1) \).
Determine the vectors joining point \( A \) to points \( B \), \( C \), and \( D \):
\( \vec{AB} = (3 - (-1))\hat{i} + (\lambda - 4)\hat{j} + (-5 - (-3))\hat{k} = 4\hat{i} + (\lambda - 4)\hat{j} - 2\hat{k} \)
\( \vec{AC} = (-3 - (-1))\hat{i} + (8 - 4)\hat{j} + (-5 - (-3))\hat{k} = -2\hat{i} + 4\hat{j} - 2\hat{k} \)
\( \vec{AD} = (-3 - (-1))\hat{i} + (2 - 4)\hat{j} + (1 - (-3))\hat{k} = -2\hat{i} - 2\hat{j} + 4\hat{k} \).
Four points are coplanar when \( [\vec{AB}\;\vec{AC}\;\vec{AD}] = 0 \):
\( \begin{vmatrix} 4 & \lambda - 4 & -2 \\ -2 & 4 & -2 \\ -2 & -2 & 4 \end{vmatrix} = 0 \).
Expanding across the first row:
\( 4[4(4) - (-2)(-2)] - (\lambda - 4)[(-2)(4) - (-2)(-2)] - 2[(-2)(-2) - 4(-2)] = 0 \)

\implies \( 4(16 - 4) - (\lambda - 4)(-8 - 4) - 2(4 + 8) = 0 \)

\implies \( 4(12) - (\lambda - 4)(-12) - 2(12) = 0 \).
Dividing by 12 throughout:
\( 4 + (\lambda - 4) - 2 = 0 \)

\implies \( \lambda + 2 - 4 = 0 \)

\implies \( \lambda - 2 = 0 
\implies \lambda = 2 \).
Thus, the required value of \( \lambda \) is 2.
In simple words: Form three direction vectors from point A, set their determinant to zero, and divide through by 12 to find lambda = 2.

Exam Tip: Notice that 12 is a common factor across all three expansion terms; factoring it out immediately avoids large arithmetic computations.

Question 9. Show that if the vectors \( \vec{a} \), \( \vec{b} \) and \( \vec{c} \) are co-planar, then \( \vec{a} + \vec{b} \), \( \vec{b} + \vec{c} \) and \( \vec{c} + \vec{a} \) are also co-planar.
Answer: Consider the scalar triple product of the vectors \( \vec{a} + \vec{b} \), \( \vec{b} + \vec{c} \), and \( \vec{c} + \vec{a} \):
\( [\vec{a} + \vec{b}\;\;\vec{b} + \vec{c}\;\;\vec{c} + \vec{a}] = 2 [\vec{a}\;\vec{b}\;\vec{c}] \).
We are given that the original vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are coplanar.
According to the coplanarity condition, their scalar triple product is zero:
\( [\vec{a}\;\vec{b}\;\vec{c}] = 0 \).
Substituting this into the identity gives:
\( [\vec{a} + \vec{b}\;\;\vec{b} + \vec{c}\;\;\vec{c} + \vec{a}] = 2(0) = 0 \).
Because the scalar triple product of \( \vec{a} + \vec{b} \), \( \vec{b} + \vec{c} \), and \( \vec{c} + \vec{a} \) is zero, these three vectors are coplanar as well.
Hence proved.
In simple words: The box product of the pair sums is twice the box product of the original vectors, so when the original is 0, the sum vectors also give 0.

Exam Tip: Be sure to cite the identity \( [\vec{a}+\vec{b}\;\vec{b}+\vec{c}\;\vec{c}+\vec{a}] = 2[\vec{a}\;\vec{b}\;\vec{c}] \) clearly to establish the core step of your proof.

Question 10. If the vectors \( \vec{A} = a\hat{i} + \hat{j} + \hat{k} \), \( \vec{B} = \hat{i} + b\hat{j} + \hat{k} \) and \( \vec{C} = \hat{i} + \hat{j} + c\hat{k} \) are co-planar. then \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1 \), where a, b, c ≠ 1
Answer: Since the vectors \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \) are coplanar, their scalar triple product must be zero:
\( \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0 \).
Apply the elementary row transformations \( R_1 \to R_1 - R_2 \) and \( R_2 \to R_2 - R_3 \):
\( \begin{vmatrix} a - 1 & 1 - b & 0 \\ 0 & b - 1 & 1 - c \\ 1 & 1 & c \end{vmatrix} = 0 \).
Expand the determinant along the first row:
\( (a - 1)[(b - 1)c - (1 - c)(1)] - (1 - b)[0 - (1 - c)(1)] + 0 = 0 \)

\implies \( (a - 1)(b - 1)c - (a - 1)(1 - c) + (1 - b)(1 - c) = 0 \).
Rewrite \( a - 1 = -(1 - a) \) and \( b - 1 = -(1 - b) \):
\( [-(1 - a)][-(1 - b)]c - [-(1 - a)](1 - c) + (1 - b)(1 - c) = 0 \)

\implies \( (1 - a)(1 - b)c + (1 - a)(1 - c) + (1 - b)(1 - c) = 0 \).
Divide every term by \( (1 - a)(1 - b)(1 - c) \) (which is non-zero since \( a, b, c \neq 1 \)):
\( \frac{c}{1 - c} + \frac{1}{1 - b} + \frac{1}{1 - a} = 0 \).
Express \( \frac{c}{1 - c} \) as \( \frac{-(1 - c) + 1}{1 - c} = -1 + \frac{1}{1 - c} \):
\( \frac{1}{1 - a} + \frac{1}{1 - b} + \left( \frac{1}{1 - c} - 1 \right) = 0 \)

\implies \( \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 1 \).
Hence proved.
In simple words: Subtract rows in the determinant to create zeroes, expand, and divide by \( (1-a)(1-b)(1-c) \) to obtain the required sum equal to 1.

Exam Tip: The algebraic conversion \( \frac{c}{1-c} = \frac{1}{1-c} - 1 \) is the decisive step; highlight this substitution clearly.