RBSE Solutions Class 12 Maths Chapter 13 Vector Exercise 13.4

Question 1. Prove that:
(i) \( \hat{i} \cdot \hat{i} + \hat{i} \cdot (-\hat{i}) = 0 \)
(ii) \( \left[ 2\hat{i} \hat{j} \hat{k} \right] + \left[ \hat{i} \hat{k} \hat{j} \right] + \left[ \hat{k} \hat{j} 2\hat{i} \right] = -1 \)
Answer:
(i) Let's prove the given statement.
L.H.S. \( = \hat{i} \cdot \hat{i} + \hat{i} \cdot (-\hat{i}) \)
We know that \( \hat{i} \cdot \hat{i} = 1 \) and \( \hat{i} \cdot (-\hat{i}) = -(\hat{i} \cdot \hat{i}) = -1 \). The dot product of a unit vector with itself is 1.
So, L.H.S. \( = 1 + (-1) \)
\( = 1 - 1 \)
\( = 0 \)
\( = \) R.H.S.
Hence, proved.
(ii) Let's prove the given statement.
L.H.S. \( = \left[ 2\hat{i} \hat{j} \hat{k} \right] + \left[ \hat{i} \hat{k} \hat{j} \right] + \left[ \hat{k} \hat{j} 2\hat{i} \right] \)
We use the scalar triple product definition: \( \left[ \vec{a} \vec{b} \vec{c} \right] = \vec{a} \cdot (\vec{b} \times \vec{c}) \).
Also, we use the cyclic property of scalar triple products: \( \left[ \vec{a} \vec{b} \vec{c} \right] = \left[ \vec{b} \vec{c} \vec{a} \right] = \left[ \vec{c} \vec{a} \vec{b} \right] \). If any two vectors are swapped, the sign changes.
\( \left[ 2\hat{i} \hat{j} \hat{k} \right] = 2 \left[ \hat{i} \hat{j} \hat{k} \right] = 2(\hat{i} \cdot (\hat{j} \times \hat{k})) = 2(\hat{i} \cdot \hat{i}) = 2(1) = 2 \)
\( \left[ \hat{i} \hat{k} \hat{j} \right] = - \left[ \hat{i} \hat{j} \hat{k} \right] = -(\hat{i} \cdot (\hat{j} \times \hat{k})) = -(\hat{i} \cdot \hat{i}) = -1 \)
\( \left[ \hat{k} \hat{j} 2\hat{i} \right] = 2 \left[ \hat{k} \hat{j} \hat{i} \right] = -2 \left[ \hat{k} \hat{i} \hat{j} \right] = -2 \left[ \hat{i} \hat{j} \hat{k} \right] = -2(1) = -2 \)
Alternatively, using the step-by-step calculation from the source:
L.H.S. \( = 2\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{i} \cdot (\hat{k} \times \hat{j}) + \hat{k} \cdot (\hat{j} \times 2\hat{i}) \)
Knowing that \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{j} = -\hat{i} \), and \( \hat{j} \times \hat{i} = -\hat{k} \), so \( \hat{j} \times 2\hat{i} = 2(\hat{j} \times \hat{i}) = -2\hat{k} \). A useful property is that `\( \hat{a} \cdot \hat{a} = 1 \)`.
L.H.S. \( = 2\hat{i} \cdot \hat{i} + \hat{i} \cdot (-\hat{i}) + \hat{k} \cdot (-2\hat{k}) \)
\( = 2(1) + (-1) + (-2)(1) \)
\( = 2 - 1 - 2 \)
\( = -1 \)
\( = \) R.H.S.
Hence, proved.
In simple words: For part (i), we used the basic rule that a unit vector dotted with itself is 1. For part (ii), we used the properties of the scalar triple product and cross products of unit vectors.

🎯 Exam Tip: Remember the cyclic property of scalar triple products \( \left[ \vec{a} \vec{b} \vec{c} \right] \) and the values of cross products for unit vectors (\( \hat{i} \times \hat{j} = \hat{k} \), etc.) to quickly solve these problems.

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 find \( \left[ \vec{a} \vec{b} \vec{c} \right] \).
Answer: To find the scalar triple product \( \left[ \vec{a} \vec{b} \vec{c} \right] \), we calculate the determinant of the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) from the three vectors.
Given vectors:
\( \vec{a} = 2\hat{i}-3\hat{j}+4\hat{k} \)
\( \vec{b} = \hat{i}+2\hat{j}-\hat{k} \)
\( \vec{c} = 3\hat{i}-\hat{j}+2\hat{k} \)
Then, \( \left[ \vec{a} \vec{b} \vec{c} \right] = \begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} \)
Let's expand the determinant along the first row:
\( = 2 \begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} - (-3) \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} + 4 \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} \)
\( = 2((2)(2) - (-1)(-1)) + 3((1)(2) - (-1)(3)) + 4((1)(-1) - (2)(3)) \)
\( = 2(4 - 1) + 3(2 + 3) + 4(-1 - 6) \)
\( = 2(3) + 3(5) + 4(-7) \)
\( = 6 + 15 - 28 \)
\( = 21 - 28 \)
\( = -7 \)
Therefore, \( \left[ \vec{a} \vec{b} \vec{c} \right] = -7 \). This value gives us information about the orientation and volume formed by the vectors.
In simple words: We put the numbers from the vectors into a square grid called a determinant. Then, we solved this determinant following a specific calculation rule to get the final answer, which is -7.

🎯 Exam Tip: Carefully compute the determinant, paying close attention to signs, especially when subtracting products in the \( 2 \times 2 \) sub-determinants. A small error can change the entire result.

Question 3. Prove that vectors \( \vec{a} = -2\hat{i}-2\hat{j}+4\hat{k} \), \( \vec{b} = -2\hat{i}+4\hat{j}-2\hat{k} \) and \( \vec{c} = 4\hat{i}-2\hat{j}-2\hat{k} \) are coplanar.
Answer: To prove that three vectors are coplanar, we need to show that their scalar triple product is zero. The scalar triple product \( \left[ \vec{a} \vec{b} \vec{c} \right] \) represents the volume of the parallelepiped formed by the three vectors. If the volume is zero, it means the vectors lie on the same plane.
Given vectors:
\( \vec{a} = -2\hat{i}-2\hat{j}+4\hat{k} \)
\( \vec{b} = -2\hat{i}+4\hat{j}-2\hat{k} \)
\( \vec{c} = 4\hat{i}-2\hat{j}-2\hat{k} \)
The scalar triple product is \( \left[ \vec{a} \vec{b} \vec{c} \right] = \begin{vmatrix} -2 & -2 & 4 \\ -2 & 4 & -2 \\ 4 & -2 & -2 \end{vmatrix} \)
Let's expand the determinant along the first row:
\( = -2 \begin{vmatrix} 4 & -2 \\ -2 & -2 \end{vmatrix} - (-2) \begin{vmatrix} -2 & -2 \\ 4 & -2 \end{vmatrix} + 4 \begin{vmatrix} -2 & 4 \\ 4 & -2 \end{vmatrix} \)
\( = -2((4)(-2) - (-2)(-2)) + 2((-2)(-2) - (-2)(4)) + 4((-2)(-2) - (4)(4)) \)
\( = -2(-8 - 4) + 2(4 + 8) + 4(4 - 16) \)
\( = -2(-12) + 2(12) + 4(-12) \)
\( = 24 + 24 - 48 \)
\( = 48 - 48 \)
\( = 0 \)
Since the scalar triple product \( \left[ \vec{a} \vec{b} \vec{c} \right] = 0 \), the given vectors \( \vec{a}, \vec{b}, \vec{c} \) are coplanar. This is the condition for coplanarity.
In simple words: We calculated a special number (determinant) using the coefficients of the three vectors. If this number turns out to be zero, it means all three vectors lie flat on the same surface, so they are coplanar.

🎯 Exam Tip: Remember that three vectors are coplanar if and only if their scalar triple product is zero. This is a key condition to apply.

Question 4. For which value of \( \lambda \), following vectors are coplanar :
(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) \( \vec{a} = \hat{i}-\hat{j}+\hat{k} \), \( \vec{b} = 2\hat{i}+\hat{j}-\hat{k} \) and \( \vec{c} = \lambda\hat{i}-\hat{j}+\lambda\hat{k} \)
Answer:
(i) For the vectors to be coplanar, their scalar triple product must be zero.
Given vectors:
\( \vec{a} = 2\hat{i}-\hat{j}+\hat{k} \)
\( \vec{b} = \hat{i}+2\hat{j}-3\hat{k} \)
\( \vec{c} = 3\hat{i}+\lambda\hat{j}+5\hat{k} \)
Set \( \left[ \vec{a} \vec{b} \vec{c} \right] = 0 \):
\( \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & \lambda & 5 \end{vmatrix} = 0 \)
Expand the determinant along the first row:
\( 2((2)(5) - (-3)(\lambda)) - (-1)((1)(5) - (-3)(3)) + 1((1)(\lambda) - (2)(3)) = 0 \)
\( 2(10 + 3\lambda) + 1(5 + 9) + 1(\lambda - 6) = 0 \)
\( 20 + 6\lambda + 14 + \lambda - 6 = 0 \)
\( 7\lambda + 28 = 0 \)
\( 7\lambda = -28 \)
\( \lambda = \frac{-28}{7} \)
\( \implies \)
\( \lambda = -4 \)
Thus, the given vectors are coplanar when \( \lambda = -4 \). This value ensures the vectors lie on the same plane.
(ii) For the vectors to be coplanar, their scalar triple product must be zero.
Given vectors:
\( \vec{a} = \hat{i}-\hat{j}+\hat{k} \)
\( \vec{b} = 2\hat{i}+\hat{j}-\hat{k} \)
\( \vec{c} = \lambda\hat{i}-\hat{j}+\lambda\hat{k} \)
Set \( \left[ \vec{a} \vec{b} \vec{c} \right] = 0 \):
\( \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ \lambda & -1 & \lambda \end{vmatrix} = 0 \)
Expand the determinant along the first row:
\( 1((1)(\lambda) - (-1)(-1)) - (-1)((2)(\lambda) - (-1)(\lambda)) + 1((2)(-1) - (1)(\lambda)) = 0 \)
\( 1(\lambda - 1) + 1(2\lambda + \lambda) + 1(-2 - \lambda) = 0 \)
\( \lambda - 1 + 3\lambda - 2 - \lambda = 0 \)
\( 3\lambda - 3 = 0 \)
\( 3\lambda = 3 \)
\( \implies \)
\( \lambda = 1 \)
Thus, the given vectors are coplanar when \( \lambda = 1 \).
In simple words: We used the rule that vectors are coplanar if their determinant (scalar triple product) is zero. We set up the determinant with the unknown value \( \lambda \) and solved the equation to find the exact value of \( \lambda \) that makes the vectors coplanar.

🎯 Exam Tip: When solving for an unknown variable like \( \lambda \), double-check your determinant expansion and algebraic simplification to avoid calculation errors. The final \( \lambda \) value must satisfy the coplanarity condition.

Question 5. Prove that the following four points are coplanar:
(i) Points derived from vectors: \( \vec{OA} = -\hat{i}+4\hat{j}-3\hat{k} \), \( \vec{OB} = 3\hat{i}+2\hat{j}-5\hat{k} \), \( \vec{OC} = -3\hat{i}+8\hat{j}-5\hat{k} \), \( \vec{OD} = -3\hat{i}+2\hat{j}+\hat{k} \)
(ii) Given points are A(0, -1, 0), B(2, 1, -1), C(1, 1, 1) and D(3, 3, 0)
Answer: To prove that four points are coplanar, we need to show that the three vectors formed by these points (e.g., \( \vec{BA}, \vec{BC}, \vec{BD} \)) are coplanar. This means their scalar triple product must be zero. The choice of reference point for forming the vectors does not change the result.
(i) Let the position vectors of the four points be:
\( \vec{OA} = -\hat{i}+4\hat{j}-3\hat{k} \)
\( \vec{OB} = 3\hat{i}+2\hat{j}-5\hat{k} \)
\( \vec{OC} = -3\hat{i}+8\hat{j}-5\hat{k} \)
\( \vec{OD} = -3\hat{i}+2\hat{j}+\hat{k} \)
Let's form three vectors using B as the common starting point:
\( \vec{BA} = \vec{OA} - \vec{OB} \)
\( = (-\hat{i}+4\hat{j}-3\hat{k}) - (3\hat{i}+2\hat{j}-5\hat{k}) \)
\( = -\hat{i}+4\hat{j}-3\hat{k}-3\hat{i}-2\hat{j}+5\hat{k} \)
\( = -4\hat{i}+2\hat{j}+2\hat{k} \)

\( \vec{BC} = \vec{OC} - \vec{OB} \)
\( = (-3\hat{i}+8\hat{j}-5\hat{k}) - (3\hat{i}+2\hat{j}-5\hat{k}) \)
\( = -3\hat{i}+8\hat{j}-5\hat{k}-3\hat{i}-2\hat{j}+5\hat{k} \)
\( = -6\hat{i}+6\hat{j}+0\hat{k} \)

\( \vec{BD} = \vec{OD} - \vec{OB} \)
\( = (-3\hat{i}+2\hat{j}+\hat{k}) - (3\hat{i}+2\hat{j}-5\hat{k}) \)
\( = -3\hat{i}+2\hat{j}+\hat{k}-3\hat{i}-2\hat{j}+5\hat{k} \)
\( = -6\hat{i}+0\hat{j}+6\hat{k} \)
Now, we calculate the scalar triple product \( \left[ \vec{BA} \vec{BC} \vec{BD} \right] \):
\( \left[ \vec{BA} \vec{BC} \vec{BD} \right] = \begin{vmatrix} -4 & 2 & 2 \\ -6 & 6 & 0 \\ -6 & 0 & 6 \end{vmatrix} \)
Expand the determinant along the first row:
\( = -4((6)(6) - (0)(0)) - 2((-6)(6) - (0)(-6)) + 2((-6)(0) - (6)(-6)) \)
\( = -4(36 - 0) - 2(-36 - 0) + 2(0 + 36) \)
\( = -4(36) - 2(-36) + 2(36) \)
\( = -144 + 72 + 72 \)
\( = -144 + 144 \)
\( = 0 \)
Since \( \left[ \vec{BA} \vec{BC} \vec{BD} \right] = 0 \), the vectors \( \vec{BA}, \vec{BC}, \vec{BD} \) are coplanar. Therefore, the four points A, B, C, D are coplanar.

(ii) Let the position vectors of the four points be:
A(0, -1, 0) \( \implies \vec{OA} = 0\hat{i}-\hat{j}+0\hat{k} \)
B(2, 1, -1) \( \implies \vec{OB} = 2\hat{i}+\hat{j}-\hat{k} \)
C(1, 1, 1) \( \implies \vec{OC} = \hat{i}+\hat{j}+\hat{k} \)
D(3, 3, 0) \( \implies \vec{OD} = 3\hat{i}+3\hat{j}+0\hat{k} \)
Let's form three vectors using B as the common starting point:
\( \vec{BA} = \vec{OA} - \vec{OB} \)
\( = (0\hat{i}-\hat{j}+0\hat{k}) - (2\hat{i}+\hat{j}-\hat{k}) \)
\( = 0\hat{i}-\hat{j}+0\hat{k}-2\hat{i}-\hat{j}+\hat{k} \)
\( = -2\hat{i}-2\hat{j}+\hat{k} \)

\( \vec{BC} = \vec{OC} - \vec{OB} \)
\( = (\hat{i}+\hat{j}+\hat{k}) - (2\hat{i}+\hat{j}-\hat{k}) \)
\( = \hat{i}+\hat{j}+\hat{k}-2\hat{i}-\hat{j}+\hat{k} \)
\( = -\hat{i}+0\hat{j}+2\hat{k} \)

\( \vec{BD} = \vec{OD} - \vec{OB} \)
\( = (3\hat{i}+3\hat{j}+0\hat{k}) - (2\hat{i}+\hat{j}-\hat{k}) \)
\( = 3\hat{i}+3\hat{j}+0\hat{k}-2\hat{i}-\hat{j}+\hat{k} \)
\( = \hat{i}+2\hat{j}+\hat{k} \)
Now, we calculate the scalar triple product \( \left[ \vec{BA} \vec{BC} \vec{BD} \right] \):
\( \left[ \vec{BA} \vec{BC} \vec{BD} \right] = \begin{vmatrix} -2 & -2 & 1 \\ -1 & 0 & 2 \\ 1 & 2 & 1 \end{vmatrix} \)
Expand the determinant along the first row:
\( = -2((0)(1) - (2)(2)) - (-2)((-1)(1) - (2)(1)) + 1((-1)(2) - (0)(1)) \)
\( = -2(0 - 4) + 2(-1 - 2) + 1(-2 - 0) \)
\( = -2(-4) + 2(-3) + 1(-2) \)
\( = 8 - 6 - 2 \)
\( = 8 - 8 \)
\( = 0 \)
Since \( \left[ \vec{BA} \vec{BC} \vec{BD} \right] = 0 \), the vectors \( \vec{BA}, \vec{BC}, \vec{BD} \) are coplanar. Thus, the four points A, B, C, D are coplanar. Calculating the determinant to be zero is the definitive proof.
In simple words: To show that four points are on the same flat surface, we pick one point and make three vectors from it to the other three points. If these three new vectors are coplanar (meaning their special "volume" calculation, the scalar triple product, is zero), then all four original points lie on the same plane. We did this for both sets of points.

🎯 Exam Tip: Always choose one point as the origin for the three vectors when proving four points are coplanar. The scalar triple product of these three vectors must be zero for coplanarity.

Question 6. Prove that \( \vec{a} = 2\hat{i}-\hat{j}+\hat{k} \), \( \vec{b} = \hat{i}-3\hat{j}-5\hat{k} \) and \( \vec{c} = 3\hat{i}-4\hat{j}-4\hat{k} \) are vector sides of right angled triangle.
Answer: For three vectors to form the sides of a right-angled triangle, they must satisfy two conditions: first, the sum of two vectors must equal the third vector (for a closed triangle), and second, the Pythagorean theorem must hold for their magnitudes (for a right angle). This means the square of the magnitude of the longest side must equal the sum of the squares of the magnitudes of the other two sides.
Let's calculate the magnitude of each vector:
Magnitude of \( \vec{a} \): \( \left| \vec{a} \right| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4+1+1} = \sqrt{6} \)
Magnitude of \( \vec{b} \): \( \left| \vec{b} \right| = \sqrt{(1)^2 + (-3)^2 + (-5)^2} = \sqrt{1+9+25} = \sqrt{35} \)
Magnitude of \( \vec{c} \): \( \left| \vec{c} \right| = \sqrt{(3)^2 + (-4)^2 + (-4)^2} = \sqrt{9+16+16} = \sqrt{41} \)
Now, let's check the Pythagorean theorem. We see that \( \left| \vec{c} \right| \) is the largest magnitude. We check if \( \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 = \left| \vec{c} \right|^2 \):
\( \left| \vec{a} \right|^2 = (\sqrt{6})^2 = 6 \)
\( \left| \vec{b} \right|^2 = (\sqrt{35})^2 = 35 \)
\( \left| \vec{c} \right|^2 = (\sqrt{41})^2 = 41 \)
Adding the squares of the two smaller magnitudes:
\( \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 = 6 + 35 = 41 \)
Since \( \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 = \left| \vec{c} \right|^2 \), the vectors satisfy the Pythagorean theorem. Additionally, we check if \( \vec{a} + \vec{b} = \vec{c} \) (or any other sum for triangle closure):
\( \vec{a} + \vec{b} = (2\hat{i}-\hat{j}+\hat{k}) + (\hat{i}-3\hat{j}-5\hat{k}) = (2+1)\hat{i} + (-1-3)\hat{j} + (1-5)\hat{k} = 3\hat{i}-4\hat{j}-4\hat{k} \).
This is exactly equal to \( \vec{c} \). Therefore, the three vectors form a right-angled triangle. This confirms they can form a closed triangle and have the angle property.
In simple words: We first found the length of each vector. Then, we checked if the square of the longest length was equal to the sum of the squares of the other two lengths (Pythagorean theorem). Since it was, and the vectors also added up correctly, they form a right-angled triangle.

🎯 Exam Tip: For vectors to form a triangle, the sum of any two vectors must be equal to or greater than the third vector in magnitude. For a right-angled triangle, the Pythagorean theorem \( |\vec{a}|^2 + |\vec{b}|^2 = |\vec{c}|^2 \) must hold, where \( |\vec{c}| \) is the largest magnitude.

Question 7. Find the volume of parallelopiped, whose coterminous edges are given by following vectors :
(i) \( \vec{a} = 4\hat{i}-3\hat{j}+\hat{k} \), \( \vec{b} = 3\hat{i}+2\hat{j}-\hat{k} \) and \( \vec{c} = 3\hat{i}-\hat{j}+2\hat{k} \)
(ii) \( \vec{a} = 2\hat{i}-3\hat{j}+\hat{k} \), \( \vec{b} = \hat{i}-\hat{j}+2\hat{k} \) and \( \vec{c} = 2\hat{i}+\hat{j}-\hat{k} \)
Answer: The volume of a parallelepiped with coterminous edges \( \vec{a}, \vec{b}, \vec{c} \) is given by the absolute value of their scalar triple product, i.e., \( \text{Volume} = \left| \left[ \vec{a} \vec{b} \vec{c} \right] \right| \). This represents the space enclosed by the vectors.
(i) Given vectors:
\( \vec{a} = 4\hat{i}-3\hat{j}+\hat{k} \)
\( \vec{b} = 3\hat{i}+2\hat{j}-\hat{k} \)
\( \vec{c} = 3\hat{i}-\hat{j}+2\hat{k} \)
First, calculate the scalar triple product \( \left[ \vec{a} \vec{b} \vec{c} \right] \):
\( \left[ \vec{a} \vec{b} \vec{c} \right] = \begin{vmatrix} 4 & -3 & 1 \\ 3 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} \)
Expand the determinant along the first row:
\( = 4((2)(2) - (-1)(-1)) - (-3)((3)(2) - (-1)(3)) + 1((3)(-1) - (2)(3)) \)
\( = 4(4 - 1) + 3(6 + 3) + 1(-3 - 6) \)
\( = 4(3) + 3(9) + 1(-9) \)
\( = 12 + 27 - 9 \)
\( = 30 \)
The volume of the parallelepiped is \( \left| 30 \right| = 30 \) cubic units. A positive volume indicates a right-handed system of vectors.
(ii) Given vectors:
\( \vec{a} = 2\hat{i}-3\hat{j}+\hat{k} \)
\( \vec{b} = \hat{i}-\hat{j}+2\hat{k} \)
\( \vec{c} = 2\hat{i}+\hat{j}-\hat{k} \)
First, calculate the scalar triple product \( \left[ \vec{a} \vec{b} \vec{c} \right] \):
\( \left[ \vec{a} \vec{b} \vec{c} \right] = \begin{vmatrix} 2 & -3 & 1 \\ 1 & -1 & 2 \\ 2 & 1 & -1 \end{vmatrix} \)
Expand the determinant along the first row:
\( = 2((-1)(-1) - (2)(1)) - (-3)((1)(-1) - (2)(2)) + 1((1)(1) - (-1)(2)) \)
\( = 2(1 - 2) + 3(-1 - 4) + 1(1 + 2) \)
\( = 2(-1) + 3(-5) + 1(3) \)
\( = -2 - 15 + 3 \)
\( = -14 \)
The volume of the parallelepiped is \( \left| -14 \right| = 14 \) cubic units. The absolute value is taken because volume cannot be negative.
In simple words: We found the volume of the 3D shape (parallelepiped) created by the three vectors. To do this, we set up and solved a determinant using the numbers from the vectors. The final answer is always a positive value, as volume cannot be negative.

🎯 Exam Tip: The volume of a parallelepiped is always a positive quantity, so remember to take the absolute value of the scalar triple product. Pay close attention to the order of vectors as changing it can flip the sign of the determinant.