Scalar triple product of vectors a, b, and c is represented as, [Tex]\vec{a}.(\vec{b}×\vec{c})[/Tex]. To calculate the scalar triple product, you can use the determinant of a 3×3 matrix formed by the components of the vectors.

Understanding the scalar triple product formula is essential for anyone studying vector calculus, as it is widely used in physics and engineering. The scalar product of three vectors, also known as the scalar triple product, is a valuable tool for calculating the volume of a parallelepiped and determining the coplanarity of vectors. In this guide, we will introduce the scalar triple product formula, explain its significance, and provide scalar triple product practice problems to help reinforce your knowledge and skills. Solve these practice questions on Scalar Triple Product to master the concept and improve your proficiency in vector calculus.

Scalar Triple Product Formula

If a = [Tex]\begin{bmatrix}a_{1}\\ a_{2}\\ a_{3}\end{bmatrix}[/Tex], b = [Tex]\begin{bmatrix}b_{1}\\ b_{2}\\ b_{3}\end{bmatrix}[/Tex], and c = [Tex]\begin{bmatrix}c_{1}\\ c_{2}\\ c_{3}\end{bmatrix}[/Tex] then,

[Tex]\vec{a}.(\vec{b}×\vec{c})[/Tex] = [Tex]\begin{vmatrix}a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3}\end{vmatrix}[/Tex]

For the given vector,

• a = a₁i + a₂j + a₃k
• b = b₁i + b₂j + b₃k
• c = c₁i + c₂j + c₃k

Scalar Triple Product Formula is,

a·(b × c) = [Tex]\begin{vmatrix} a_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{vmatrix}[/Tex]

Scalar Triple Product Practice Problems: Solved

Problem 1: Given vectors a = (1, 2, 3), b = (4, 5, 6), and c = (7, 8, 9), calculate a·(b × c).

Solution:

Step 1: Set up the determinant for the scalar triple product.

Scalar triple product a·(b × c) can be calculated using the determinant: [Tex]\begin{vmatrix} a_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{vmatrix}[/Tex]

Step 2: Substitute the values from the given vectors.

[Tex]\begin{vmatrix} 1& 4& 7\\ 2& 5& 8\\ 3& 6& 9\end{vmatrix}[/Tex]

Step 3: Calculate the determinant using the following method:

(a₁b₂c₃ + a₂b₃c₁ + a₃b₁c₂) – (a₃b₂c₁ + a₁b₃c₂ + a₂b₁c₃)

= (1×5×9 + 2×6×7 + 3×4×8) – (3×5×7 + 1×6×8 + 2×4×9)

Step 4: Simplify the terms.

= (45 + 84 + 96) – (105 + 48 + 72)

= 225 – 225

= 0

Therefore, the scalar triple product a·(b × c) = 0.

Problem 2: Prove that a · (b × c) = -c · (b × a) for any vectors a, b, and c.

Solution:

First, recall the definition of the scalar triple product:

a · (b × c) = [a b c]

Where [a b c] represents the determinant of the matrix formed by vectors a, b, and c as columns.

Now, let’s consider the right side of the equation we want to prove:

c · (b × a)

We can rewrite this using the scalar triple product notation:c · (b × a) = –

A property of determinants is that swapping any two columns changes the sign of the determinant. So: = -[a b c]

Therefore: = [a b c]

Combining steps 3 and 5:

c · (b × a) = [a b c]

But we know from step 1 that:

[a b c] = a · (b × c)

Therefore:

-c · (b × a) = a · (b × c)

Thus, we have proven that a · (b × c) = -c · (b × a) for any vectors a, b, and c.

Problem 3: If a = 2i + 3j – k, b = i – 2j + 4k, and c = -i + j + 2k, calculate the volume of the parallelepiped formed by these vectors.

Solution:

Step 1: First, let’s recall the formula for the volume of a parallelepiped:

Volume = |a · (b × c)|

where × denotes cross product and · denotes dot product.

Let’s start by calculating b × c:

b × c = (i – 2j + 4k) × (-i + j + 2k)

= [Tex]\begin{vmatrix} i& j& k\\ 1& -2& 4\\ -1& 1& 2\end{vmatrix}[/Tex]

= {(-2)(2) – (4)(1)}i + {(1)(2) – (4)(-1)}j + {(1)(1) – (-2)(-1)}k

= -4i + 6j – 1k

Step 2: Now we have a · (b × c):

a · (b × c) = (2i + 3j – k) · (-4i + 6j – k)

= 2(-4) + 3(6) + (-1)(-1)

= -8 + 18 + 1

= 11

Volume is the absolute value of this scalar triple product = |11| = 11

Therefore, the volume of the parallelepiped formed by vectors a, b, and c is 11 cubic units.

Problem 4: Prove that a · (a × b) = 0 for any vectors a and b.

Solution:

Let’s start by defining vectors a and b in 3D space:

a = (a₁, a₂, a₃)

b = (b₁, b₂, b₃)

Now, let’s calculate a × b:

a × b = (a₂b₃ – a₃b₂, a₃b₁ – a₁b₃, a₁b₂ – a₂b₁)

Next, we’ll perform the dot product of a with (a × b):

a · (a × b) = a₁(a₂b₃ – a₃b₂) + a₂(a₃b₁ – a₁b₃) + a₃(a₁b₂ – a₂b₁)

Let’s expand this expression:

a · (a × b) = a₁a₂b₃ – a₁a₃b₂ + a₂a₃b₁ – a₂a₁b₃ + a₃a₁b₂ – a₃a₂b₁

Now, let’s rearrange the terms:

a · (a × b) = (a₁a₂b₃ – a₂a₁b₃) + (a₂a₃b₁ – a₃a₂b₁) + (a₃a₁b₂ – a₁a₃b₂)

We can factor out common terms:

a · (a × b) = (a₁a₂ – a₂a₁)b₃ + (a₂a₃ – a₃a₂)b₁ + (a₃a₁ – a₁a₃)b₂

Notice that each parenthesis contains two terms that are negatives of each other:

a₁a₂ – a₂a₁ = 0

a₂a₃ – a₃a₂ = 0

a₃a₁ – a₁a₃ = 0

Therefore, each term in the expression becomes zero:

a · (a × b) = 0 · b₃ + 0 · b₁ + 0 · b₂ = 0

Thus, we have proven that a · (a × b) = 0 for any vectors a and b.

Problem 5: Given that a · (b × c) = 10, b · (c × a) = 20, and c · (a × b) = 30, find the value of a · (c × b).

Solution:

First, recall the cyclic property of scalar triple products:

a · (b × c) = b · (c × a) = c · (a × b)

From the given information, we can see that this property doesn’t hold for the given values. This suggests that the vectors a, b, and c form a left-handed system.

Now, let’s consider the relationship between (b × c) and (c × b):

b × c = -(c × b)

This means that:

a · (b × c) = -a · (c × b)

We’re given that a · (b × c) = 10, so:

a · (c × b) = -10

Therefore, the value of a · (c × b) is -10.

Problem 6: Given vectors a = (2, -1, 3), b = (1, 2, -2), and c = (3, 0, 1), determine if these vectors are coplanar.

Solution:

Recall that the scalar triple product is given by a · (b × c).

First, let’s calculate b × c:

b × c = [Tex]\begin{vmatrix} i& j& k\\ 1& 2& -2\\ 3& 0& 1\end{vmatrix}[/Tex]

= (2(1) – (-2)(0))i – (1(1) – (-2)(3))j + (1(0) – 2(3))k

= 2i – (1 + 6)j – 6k

= 2i – 7j – 6k

Now, let’s calculate a · (b × c):

a · (b × c) = (2, -1, 3) · (2, -7, -6)

= 2(2) + (-1)(-7) + 3(-6)

= 4 + 7 – 18

= -7

Since the scalar triple product is not zero (it’s -7), these vectors are not coplanar.

Therefore, vectors a, b, and c are not coplanar.

Problem 7: If a · (b × c) = 6 and |a| = 2, |b| = 3, |c| = 1, find the angle θ between b and c.

Solution:

Step 1: Recall the formula for scalar triple product

|a · (b × c)| = |a||b||c|sin(θ)

Step 2: Substitute known values

6 = 2 × 3 × 1 × sin(θ)

Step 3: Solve for sin(θ)

6 = 6sin(θ)

sin(θ) = 1

Step 4: Find θ

θ = arcsin(1) = 90°

Therefore, the angle between b and c is 90°.

Problem 8: Prove that (a × b) · (c × d) = (a · c)(b · d) – (a · d)(b · c)

Solution:

Step 1: Express the left side using the scalar triple product

(a × b) · (c × d) = c · (d × (a × b))

Step 2: Use the vector triple product identity

c · (d × (a × b)) = c · ((d · b)a – (d · a)b)

Step 3: Distribute the dot product

= (d · b)(c · a) – (d · a)(c · b)

Step 4: Rearrange terms

= (a · c)(b · d) – (a · d)(b · c)

This proves the identity.

Problem 9: Given a · (b × c) = 10 and d = 2a + 3b – c, calculate d · (b × c).

Solution:

Step 1: Use the distributive property of scalar triple product

d · (b × c) = (2a + 3b – c) · (b × c)

= 2(a · (b × c)) + 3(b · (b × c)) – (c · (b × c))

Step 2: Simplify using known properties

a · (b × c) = 10 (given)

b · (b × c) = 0 (a vector dotted with its own cross product is always zero)

c · (b × c) = 0 (same reason as above)

Step 3: Substitute and calculate

d · (b × c) = 2(10) + 3(0) – (0)

= 20

Therefore, d · (b × c) = 20.

Problem 10: If a · (b × c) = 5, find (2a) · ((3b) × (-c)).

Solution:

Step 1: Use the scalar multiplication property of scalar triple product

(2a) · ((3b) × (-c)) = 2 × 3 × (-1) × (a · (b × c))

Step 2: Simplify

= -6 × (a · (b × c))

Step 3: Substitute the given value

= -6 × 5

= -30

Therefore, (2a) · ((3b) × (-c)) = -30

Practice Questions Scalar Triple Product: Unsolved

Q1. Calculate the scalar triple product of a = (1, 2, 3), b = (4, 5, 6), and c = (7, 8, 9).

Q2. Determine if the following vectors are coplanar: a = (1, 0, 2), b = (3, 1, 0), c = (2, 1, 1)

Q3. Find the volume of the parallelepiped formed by the vectors: a = (2, 0, -1), b = (1, 2, 3), c = (0, 1, -2)

Q4. If a · (b × c) = 10 and |a| = 2, |b| = 3, |c| = 1, find the angle between a and (b × c).

Q5. Prove that a · (b × c) = b · (c × a) = c · (a × b)

Q6. Given a = i + 2j – k, b = 2i – j + 3k, and c = 3i + 4j – 2k, calculate a · (b × c).

Q7. If a · (b × c) = 0, what can you conclude about vectors a, b, and c?

Q8. Calculate the scalar triple product of a = (x, y, z), b = (1, 1, 1), and c = (x², y², z²).

Q9. Show that the scalar triple product changes sign when any two vectors are interchanged.

Q10. If a, b, and c are unit vectors such that a is perpendicular to both b and c, and the angle between b and c is 60°, find a · (b × c).

Scalar Triple Product Practice Problems- FAQs

What Does Scalar Triple Product Represent Geometrically?

It represents the volume of the parallelepiped formed by three vectors.

When is the Scalar Triple Product Equal to Zero?

When the three vectors are coplanar or when any two vectors are parallel.

What is the Cyclic Permutation Property of Scalar Triple Product?

a · (b × c) = b · (c × a) = c · (a × b)

How does Swapping Two Vectors Affect the Scalar Triple Product?

It changes the sign of the result.

Can the Scalar Triple Product be Expressed as a Determinant?

Yes, it can be expressed as a 3×3 determinant of the vectors’ components.