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JEE • MATHEMATICS • SCALAR TRIPLE PRODUCT

Scalar triple product practice · Easy

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181 Easy

181 of 185

Application +4 | -1

The position vectors of three vertices A, B, C of a tetrahedron OABC with respect to its vertex O are

\hat i,\,6\hat j,\,\hat k,\,

then its volume (in cu units) is

View answer and explanation

Correct option: D

Option D (6) is correct because the volume of the tetrahedron is calculated using the scalar triple product of the position vectors divided by 6.

Solution approach

Correct Answer + Core Reason: 6. The volume of a tetrahedron is calculated using the scalar triple product and dividing by 6.

Step-by-Step Solution:

Step 1: Given $A = (6, 0, 0), B = (0, 6, 0), C = (0, 0, 1)$ and $O = (0, 0, 0)$ .

Step 2: $OA = 6\widehat i + 0\widehat j + 0\widehat k, OB = 0\widehat i + 6\widehat j + 0\widehat k, OC = 0\widehat i + 0\widehat j + \widehat k$

Step 3: Volume of tetrahedron $= \frac{1}{6}[OA \, OB \, OC] = \frac{1}{6}[6\widehat i \, 6\widehat j \, \widehat k]$

Step 4: Calculate the determinant: $\frac{1}{6}\left| {\begin{{array}}{{{*{{20}}{l}}}} 6&0&0 \\ 0&6&0 \\ 0&0&1 \end{{array}}} \right| = \frac{1}{6}[6(6 - 0)] = 6$

Step 5: Volume = 6 cubic units.

Why other options are wrong:
- Option A: This option is incorrect because it is not the correct volume calculated using the scalar triple product.
- Option B: This option is incorrect and doesn't relate to the calculation.
- Option C: This option is incorrect and doesn't relate to the calculation.

Key Concepts:
- Volume of a tetrahedron
- Scalar triple product
- Determinants

Memory Aids:
- Remember the formula for the volume of a tetrahedron: $V = \frac{1}{6}|[a \, b \, c]|$ where a, b, and c are vectors representing three edges meeting at a vertex.

Quick Tips:
- Always remember to divide by 6 when calculating the volume of a tetrahedron using the scalar triple product.

182 Easy

182 of 185

Application +4 | -1

The three concurrent edges of a parallelopiped represent the vectors

\vec{a}, \vec{b}, \vec{c}

such that

\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=V

. Then, the volume of the parallelopiped whose three concurrent edges are the three diagonals of three faces of the given parallelopiped is

View answer and explanation

Correct option: A

2 \mathrm{~V}

Option A (

2 \mathrm{~V}

) is correct because the volume of the new parallelepiped is twice the volume of the original parallelepiped.

Solution approach

Correct Answer + Core Reason: $2 \mathrm{~V}$ . The volume of the new parallelopiped is twice the original volume.

Step-by-Step Solution:

Step 1: Define the edges of the new parallelepiped: $\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}$ .

Step 2: Calculate the volume of the new parallelepiped: $\left[\vec a + \vec b\,\,\vec b + \vec c\,\,\vec c + \vec a\right]$ .

Step 3: Expand the scalar triple product: $\left[\vec a + \vec b\,\,\vec b + \vec c\,\,\vec c + \vec a\right] = ((\vec a + \vec b) \times (\vec b + \vec c)) \cdot (\vec c + \vec a)$ .

Step 4: Simplify the cross product: $(\vec a + \vec b) \times (\vec b + \vec c) = \vec a \times \vec b + \vec a \times \vec c + \vec b \times \vec b + \vec b \times \vec c = \vec a \times \vec b + \vec a \times \vec c + \vec b \times \vec c$ .

Step 5: Take the dot product with $(\vec c + \vec a)$ : $(\vec a \times \vec b + \vec a \times \vec c + \vec b \times \vec c) \cdot (\vec c + \vec a) = (\vec a \times \vec b) \cdot \vec c + (\vec a \times \vec b) \cdot \vec a + (\vec a \times \vec c) \cdot \vec c + (\vec a \times \vec c) \cdot \vec a + (\vec b \times \vec c) \cdot \vec c + (\vec b \times \vec c) \cdot \vec a$ .

Step 6: Simplify using the properties of the scalar triple product: $\left[ {\begin{array}{llllllllllllllllllll} {\vec a}&{\vec b}&{\vec c} \end{array}} \right] + 0 + 0 + \left[ {\begin{array}{llllllllllllllllllll} {\vec b}&{\vec c}&{\vec a} \end{array}} \right] = 2\left[ {\begin{array}{llllllllllllllllllll} {\vec a}&{\vec b}&{\vec c} \end{array}} \right] = 2V$ .

Why other options are wrong:
- Option B: This is incorrect because the volume is twice the original volume.
- Option C: This is incorrect because the volume is twice the original volume.
- Option D: This is incorrect because the volume is twice the original volume.

Key Concepts:
- Scalar triple product
- Volume of a parallelepiped
- Properties of dot and cross products

Memory Aids:
- Remember the properties of scalar triple products and how they relate to volume.

Quick Tips:
- Use the properties of scalar triple products to simplify expressions.

183 Easy

183 of 185

Application +4 | -1

The three vectors

\mathbf{i}+\mathbf{j}, \mathbf{j}+\mathbf{k}, \mathbf{k}+\mathbf{i}

taken two at a time form three planes. The three unit vectors drawn perpendicular to these three planes form a parallelopiped of volume

View answer and explanation

Correct option: D

\frac{4}{3 \sqrt{3}}

Option D (

\frac{{4}}{{3 \sqrt{3}}})

is correct because the volume of the parallelepiped formed by the unit vectors is calculated using the scalar triple product, which results in

\frac{{4}}{{3 \sqrt{3}}}

Solution approach

Correct Answer + Core Reason: $\frac{{4}}{{3 \sqrt{3}}}$ . The volume of the parallelepiped is calculated using the scalar triple product of the unit vectors.

Step-by-Step Solution:

Step 1: Find the vectors perpendicular to the planes formed by the given vectors taken two at a time.\nVector 1: $\mathbf{a} = \mathbf{i} + \mathbf{j}$ \nVector 2: $\mathbf{b} = \mathbf{j} + \mathbf{k}$ \nVector 3: $\mathbf{c} = \mathbf{k} + \mathbf{i}$

Step 2: Calculate the cross products of each pair of vectors:
$\mathbf{a} \times \mathbf{b} = (\mathbf{i} + \mathbf{j}) \times (\mathbf{j} + \mathbf{k}) = \mathbf{i} - \mathbf{j} + \mathbf{k}$
$\mathbf{b} \times \mathbf{c} = (\mathbf{j} + \mathbf{k}) \times (\mathbf{k} + \mathbf{i}) = \mathbf{i} + \mathbf{j} - \mathbf{k}$
$\mathbf{c} \times \mathbf{a} = (\mathbf{k} + \mathbf{i}) \times (\mathbf{i} + \mathbf{j}) = -\mathbf{i} + \mathbf{j} + \mathbf{k}$

Step 3: Find the unit vectors in the direction of these cross products:
$\hat{\mathbf{n}}_1 = \frac{{\mathbf{i} - \mathbf{j} + \mathbf{k}}}{{\sqrt{3}}}$
$\hat{\mathbf{n}}_2 = \frac{{\mathbf{i} + \mathbf{j} - \mathbf{k}}}{{\sqrt{3}}}$
$\hat{\mathbf{n}}_3 = \frac{{-\mathbf{i} + \mathbf{j} + \mathbf{k}}}{{\sqrt{3}}}$

Step 4: Calculate the volume of the parallelepiped formed by these unit vectors using the scalar triple product:\nVolume = $|[\hat{\mathbf{n}}_1 \\, \hat{\mathbf{n}}_2 \\, \hat{\mathbf{n}}_3]$

184 Easy

184 of 185

Analyzing +4 | -1

\mathbf{u}, \mathbf{v}

and

\mathbf{w}

are three non-coplanar vectors, then

(\mathbf{u}+\mathbf{v}-\mathbf{w}) \cdot[(\mathbf{u}-\mathbf{v}) \times(\mathbf{v}-\mathbf{w})]

equal to

View answer and explanation

Correct option: B

\mathbf{u} \cdot \mathbf{v} \times \mathbf{w}

Option B (

\mathbf{u} \cdot \mathbf{v} \times \mathbf{w}

) is correct because expanding the expression and simplifying using properties of scalar triple products results in

\mathbf{u} \cdot \mathbf{v} \times \mathbf{w}

Solution approach

Correct Answer + Core Reason: $\mathbf{u} \cdot \mathbf{v} \times \mathbf{w}$ . Expanding and simplifying the expression leads to this result.

Step-by-Step Solution:

Step 1: Given $(\mathbf{u}+\mathbf{v}-\mathbf{w}) \cdot [(\mathbf{u}-\mathbf{v}) \times (\mathbf{v}-\mathbf{w})]$ .

Step 2: Expand the cross product:
$(\mathbf{u}-\mathbf{v}) \times (\mathbf{v}-\mathbf{w}) = \mathbf{u} \times \mathbf{v} - \mathbf{u} \times \mathbf{w} - \mathbf{v} \times \mathbf{v} + \mathbf{v} \times \mathbf{w} = \mathbf{u} \times \mathbf{v} - \mathbf{u} \times \mathbf{w} + \mathbf{v} \times \mathbf{w}$ .

Step 3: Substitute the expanded cross product into the original expression:
$(\mathbf{u}+\mathbf{v}-\mathbf{w}) \cdot (\mathbf{u} \times \mathbf{v} - \mathbf{u} \times \mathbf{w} + \mathbf{v} \times \mathbf{w}) = \mathbf{u} \cdot (\mathbf{u} \times \mathbf{v}) - \mathbf{u} \cdot (\mathbf{u} \times \mathbf{w}) + \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{v} \cdot (\mathbf{u} \times \mathbf{v}) - \mathbf{v} \cdot (\mathbf{u} \times \mathbf{w}) + \mathbf{v} \cdot (\mathbf{v} \times \mathbf{w}) - \mathbf{w} \cdot (\mathbf{u} \times \mathbf{v}) + \mathbf{w} \cdot (\mathbf{u} \times \mathbf{w}) - \mathbf{w} \cdot (\mathbf{v} \times \mathbf{w})$ .

Step 4: Simplify using the properties of scalar triple products:
$\mathbf{u} \cdot (\mathbf{u} \times \mathbf{v}) = 0$ , $\mathbf{u} \cdot (\mathbf{u} \times \mathbf{w}) = 0$ , $\mathbf{v} \cdot (\mathbf{u} \times \mathbf{v}) = 0$ , $\mathbf{v} \cdot (\mathbf{v} \times \mathbf{w}) = 0$ , $\mathbf{w} \cdot (\mathbf{u} \times \mathbf{w}) = 0$ , $\mathbf{w} \cdot (\mathbf{v} \times \mathbf{w}) = 0$ .\nSo, the expression becomes: $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) - \mathbf{v} \cdot (\mathbf{u} \times \mathbf{w}) - \mathbf{w} \cdot (\mathbf{u} \times \mathbf{v}) = \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) - \mathbf{w} \cdot (\mathbf{u} \times \mathbf{v})$ .

Step 5: Use the property $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b})$ :
$\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) - \mathbf{w} \cdot (\mathbf{u} \times \mathbf{v}) = \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$ .

Final Answer: $\mathbf{u} \cdot \mathbf{v} \times \mathbf{w}$ .

Why other options are wrong:
- Option A: Incorrect because the expression does not simplify to 0.
- Option C: Incorrect because the expression does not simplify to $\mathbf{u} \cdot \mathbf{w} \times \mathbf{v}$ .
- Option D: Incorrect because the expression does not simplify to $3 \mathbf{u} \cdot \mathbf{v} \times \mathbf{w}$ .

Key Concepts:
- Scalar triple product
- Vector triple product

Memory Aids:
- Remember the properties of scalar triple products: $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b})$
- Remember that $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0$ .

Quick Tips:
- Expand the expression carefully.
- Use the properties of scalar triple products to simplify.

185 Easy

185 of 185

Application +4 | -1

Let

\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}

and

\vec{c}=x \hat{i}+(x-2) \hat{j}-\hat{k}

. If the vector

\vec{c}

lies in the plane of

\vec{a}

and

\vec{b}

, then

x

equals

View answer and explanation

Correct option: B

-2

The correct answer is B (-2) because if

\vec{c}

lies in the plane of

\vec{a}

and

\vec{b}

, then the scalar triple product of the three vectors is zero.

Solution approach

Correct Answer + Core Reason: -2. The scalar triple product is zero for coplanar vectors.

Step-by-Step Solution:

Step 1: State the coplanarity condition.\nIf $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$ , then $\vec{a}, \vec{b}, \vec{c}$ are coplanar, so $[\vec{a} \,\vec{b} \,\vec{c}] = 0$ .

Step 2: Write the determinant for the scalar triple product.
$\left| {\begin{{array}}{{*{20}{c}} 1&1&1 \\ 1&{ - 1}&2 \\ x&{x - 2}&{ - 1} \end{{array}}} \right| = 0$

Step 3: Evaluate the determinant.
$1(1 - 2x + 4) - 1(-1 - 2x) + 1(x - 2 + x) = 0$
$5 - 2x + 1 + 2x + 2x - 2 = 0$
$2x + 4 = 0$

Step 4: Solve for x.
$2x = -4$
$x = -2$

Why other options are wrong:
- Option A: -4 is incorrect because it does not satisfy the coplanarity condition when substituted into the determinant.
- Option C: 0 is incorrect because it does not satisfy the coplanarity condition when substituted into the determinant.
- Option D: 1 is incorrect because it does not satisfy the coplanarity condition when substituted into the determinant.

Key Concepts:
- Coplanar vectors
- Scalar triple product
- Determinant calculation

Memory Aids:
- Coplanarity condition: $[\vec{a} \,\vec{b} \,\vec{c}] = 0$

Quick Tips:
- Remember the determinant formula for the scalar triple product.
- Double-check the signs when evaluating the determinant.

Revision notes and key takeaways

Review SCALAR TRIPLE PRODUCT in your textbook or class notes, then use this set to check your understanding.

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