The two co-initial adjacent sides of a parallelogram are 2ı̂ − 4ȷ̂ − 5k̂ and 2ı̂ + 2ȷ̂ + 3k̂ . Find its diagon

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Q. The two co-initial adjacent sides of a parallelogram are 2ı̂ - 4ȷ̂ - 5k̂ and 2ı̂ + 2ȷ̂ + 3k̂ . Find its diagonals and use them to find the area of the parallelogram.

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Concept understanding — Vector Triple Product

The Vector Triple Product

When you cross three vectors together as $\vec{a}\times(\vec{b}\times\vec{c})$ , the result is again a vector — this is the vector triple product. The remarkable thing is that it can always be rewritten without any cross products at all, using only dot products.

The Key Identity (the "BAC $-$ CAB" rule)

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

A memorable way to recall it: the answer is B times (A dot C) minus C times (A dot B) — "BAC minus CAB". The two survivors are $\vec{b}$ and $\vec{c}$ (the vectors inside the inner bracket); each is scaled by a dot product involving the outside vector $\vec{a}$ .

Why the Result Lies in the Plane of $\vec{b}$ and $\vec{c}$

The inner product $\vec{b}\times\vec{c}$ is perpendicular to the plane containing $\vec{b}$ and $\vec{c}$ . Crossing $\vec{a}$ with that perpendicular swings the result back into the $\vec{b}$ – $\vec{c}$ plane. So the answer must be a combination $\lambda\vec{b}+\mu\vec{c}$ — and the identity tells you exactly what $\lambda$ and $\mu$ are.

Order Matters — the Product Is Not Associative

The brackets are not decoration. In general,

\vec{a}\times(\vec{b}\times\vec{c}) \ne (\vec{a}\times\vec{b})\times\vec{c}.

The left side lies in the plane of $\vec{b},\vec{c}$ ; the right side lies in the plane of $\vec{a},\vec{b}$ . Its own expansion is

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

which is a different vector. Always keep the parentheses where they are given.

Watch out

A frequent slip is to "cancel" and write $\vec{a}\times(\vec{b}\times\vec{c})$ as some multiple of $\vec{a}$ . It is not — the surviving vectors are $\vec{b}$ and $\vec{c}$ , never the outer vector.

Quick Example

Let $\vec{a}=\hat{i}$ , $\vec{b}=\hat{j}$ , $\vec{c}=\hat{i}$ . Then $\vec{a}\cdot\vec{c}=1$ and $\vec{a}\cdot\vec{b}=0$ , so

\vec{a}\times(\vec{b}\times\vec{c}) = (1)\hat{j} - (0)\hat{i} = \hat{j}.

Checking directly: $\vec{b}\times\vec{c}=\hat{j}\times\hat{i}=-\hat{k}$ , and $\hat{i}\times(-\hat{k})=\hat{j}$ . The identity agrees.

The vector triple product identity goes beyond the core NCERT Class 12 Vector Algebra syllabus, but it is an important topic for JEE Advanced and select state CETs, building on the scalar and vector product foundations already laid in the NCERT curriculum. Students searching "BAC CAB rule vector triple product" should master the basic cross product and dot product first, since this identity is really just a compressed combination of both.

The diagonals of a parallelogram are $\vec a + \vec b$ and $\vec a - \vec b$ , and its area is $\tfrac{1}{2}\lvert \vec d_1 \times \vec d_2\rvert$ . Here $\vec d_1 = 4\hat\imath - 2\hat\jmath - 2\hat k$ , $\vec d_2 = -6\hat\jmath - 8\hat k$ , and the area is $\mathbf{2\sqrt{101}}$ square units.

Let the co-initial adjacent sides be

\vec a = 2\hat\imath - 4\hat\jmath - 5\hat k, \qquad \vec b = 2\hat\imath + 2\hat\jmath + 3\hat k.

Diagonals (parallelogram law):

\vec d_1 = \vec a + \vec b = 4\hat\imath - 2\hat\jmath - 2\hat k,

\vec d_2 = \vec a - \vec b = -6\hat\jmath - 8\hat k.

Area $= \tfrac{1}{2}\lvert \vec d_1 \times \vec d_2 \rvert$ , since $(\vec a + \vec b)\times(\vec a - \vec b) = -2(\vec a \times \vec b)$ .

\vec d_1 \times \vec d_2 = \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\4 & -2 & -2 \\0 & -6 & -8 \end{vmatrix} = \hat\imath(16 - 12) - \hat\jmath(-32 - 0) + \hat k(-24 - 0) = 4\hat\imath + 32\hat\jmath - 24\hat k.

\lvert \vec d_1 \times \vec d_2 \rvert = \sqrt{4^2 + 32^2 + (-24)^2} = \sqrt{16 + 1024 + 576} = \sqrt{1616} = 4\sqrt{101}.

\text{Area} = \tfrac{1}{2}(4\sqrt{101}) = 2\sqrt{101} \text{ square units}.

✓Final answer

\vec d_1 = 4\hat\imath - 2\hat\jmath - 2\hat k

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 21 on this concept.

CBSE 2026Set A1 markMCQ
$\vec{i} \cdot (\vec{j} \times \vec{k}) =$
(a) $1$
(b) $0$
(c) $-1$
(d) $\vec{i}$
›Reveal solutionSolution

$\vec{j}\times\vec{k} = \vec{i}$ , and $\vec{i}\cdot\vec{i} = 1$ .

Using the right-handed rule, $\vec{j} \times \vec{k} = \vec{i}$ . Then

$\vec{i} \cdot (\vec{j} \times \vec{k}) = \vec{i} \cdot \vec{i} = 1.$

This is the scalar triple product of the unit vectors, which equals $1$ .

✓Final answer
$1$
CBSE 2026Set A1 markMCQ
$\vec{a} \cdot (\vec{a} \times \vec{a}) =$
(a) $1$
(b) $0$
(c) $\vec{a}$
(d) $-1$
›Reveal solutionSolution

$\vec{a}\times\vec{a}=\vec{0}$ , hence $\vec{a}\cdot\vec{0}=0$ .

Any vector crossed with itself is the zero vector: $\vec{a}\times\vec{a} = \vec{0}$ . Therefore

$\vec{a}\cdot(\vec{a}\times\vec{a}) = \vec{a}\cdot\vec{0} = 0.$

✓Final answer
$0$
CBSE 2026Set ANNUAL1 markMCQ
$\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})$
(a) $0$
(b) $-1$
(c) $1$
(d) $3$
›Reveal solutionSolution

$1+(-1)+1=1$ .

$\hat i\cdot(\hat j\times\hat k)=\hat i\cdot\hat i=1$ .

$\hat j\cdot(\hat i\times\hat k)=\hat j\cdot(-\hat j)=-1$ .

$\hat k\cdot(\hat i\times\hat j)=\hat k\cdot\hat k=1$ .

Sum $=1-1+1=1$ .

✓Final answer
$1$
CBSE 2025Set E1 markMCQ
$(\vec{k} \times \vec{j}) \cdot \vec{i} =$
(a) $0$
(b) $1$
(c) $-1$
(d) $2\vec{i}$
›Reveal solutionSolution

This is a scalar triple product; its value is $-1$ .

Using the right-hand cyclic rule, $\vec{j}\times\vec{k}=\vec{i}$ , $\vec{k}\times\vec{i}=\vec{j}$ , $\vec{i}\times\vec{j}=\vec{k}$ . Reversing the order changes the sign, so

$\vec{k}\times\vec{j}=-(\vec{j}\times\vec{k})=-\vec{i}.$

Then

$(\vec{k}\times\vec{j})\cdot\vec{i}=(-\vec{i})\cdot\vec{i}=-1.$

✓Final answer
$-1$
CBSE 2025Set ANNUAL1 markMCQ
$\hat{i}\cdot(\hat{k}\times\hat{j})-\hat{j}\cdot(\hat{k}\times\hat{i})+\hat{k}\cdot(\hat{i}\times\hat{j})$
(a) $1$
(b) $0$
(c) $-3$
(d) $-1$
›Reveal solutionSolution

Evaluate each triple-scalar-product term using the standard identities $\hat{i}\times\hat{j}=\hat{k}$ , $\hat{j}\times\hat{k}=\hat{i}$ , $\hat{k}\times\hat{i}=\hat{j}$ .

$\hat{k}\times\hat{j}=-(\hat{j}\times\hat{k})=-\hat{i}$ , so $\hat{i}\cdot(\hat{k}\times\hat{j})=\hat{i}\cdot(-\hat{i})=-1$ .

$\hat{k}\times\hat{i}=\hat{j}$ , so $\hat{j}\cdot(\hat{k}\times\hat{i})=\hat{j}\cdot\hat{j}=1$ , hence $-\hat{j}\cdot(\hat{k}\times\hat{i})=-1$ .

$\hat{i}\times\hat{j}=\hat{k}$ , so $\hat{k}\cdot(\hat{i}\times\hat{j})=\hat{k}\cdot\hat{k}=1$ .

Adding: $(-1)+(-1)+1=-1$ .

✓Final answer
$-1$
CBSE 2024Set D1 markMCQ
$\vec{i} \cdot (\vec{j} \times \vec{k}) =$
(a) $1$
(b) $0$
(c) $-1$
(d) $\vec{i}$
›Reveal solutionSolution

This is the scalar triple product $[\vec{i}\ \vec{j}\ \vec{k}] = 1$ .

$\vec{j}\times\vec{k} = \vec{i}$ , so $\vec{i}\cdot(\vec{j}\times\vec{k}) = \vec{i}\cdot\vec{i} = 1.$ (It is the volume of the unit cube.)

✓Final answer
$1$
CBSE 2023Set 65/1/11 markMCQ
$(\hat{i} \times \hat{j}) \cdot \hat{j} + (\hat{j} \times \hat{i}) \cdot \hat{k}$
›Reveal solutionSolution
We evaluate the expression by applying the properties of cross and dot products for orthonormal unit vectors. The first term $(\hat{i} \times \hat{j}) \cdot \hat{j}$ simplifies to $0$ , and the second term $(\hat{j} \times \hat{i}) \cdot \hat{k}$ simplifies to $-1$ . The sum is -1.

The problem asks us to evaluate an expression involving the cross product and dot product of the standard orthonormal unit vectors $\hat{i}$ , $\hat{j}$ , and $\hat{k}$ . These vectors represent the directions along the positive x, y, and z axes, respectively, and each has a magnitude of 1.

The cross product of two vectors results in a vector perpendicular to both original vectors. For $\hat{i}$ , $\hat{j}$ , $\hat{k}$ , they follow a right-hand rule:

$\hat{i} \times \hat{j} = \hat{k}$

$\hat{j} \times \hat{k} = \hat{i}$

$\hat{k} \times \hat{i} = \hat{j}$ The cross product is anti-commutative, meaning reversing the order of the vectors changes the sign of the result: $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$ . For example, $\hat{j} \times \hat{i} = -\hat{k}$ .

The dot product of two vectors results in a scalar. It measures the extent to which two vectors point in the same direction.

If two vectors are orthogonal (perpendicular), their dot product is 0. For example, $\hat{i} \cdot \hat{j} = 0$ .

If two vectors are parallel, their dot product is the product of their magnitudes. For unit vectors, $\hat{a} \cdot \hat{a} = |\hat{a}|^2 = 1^2 = 1$ .

Let's apply these properties to evaluate the given expression term by term.

The expression we need to evaluate is $(\hat{i} \times \hat{j}) \cdot \hat{j} + (\hat{j} \times \hat{i}) \cdot \hat{k}$ .

Evaluate the first term: $(\hat{i} \times \hat{j}) \cdot \hat{j}$

First, we determine the cross product $\hat{i} \times \hat{j}$ .
$\hat{i}$
Substituting this into the first term, we get:

$(\hat{i} \times \hat{j}) \cdot \hat{j} = \hat{k} \cdot \hat{j}$

Next, we evaluate the dot product $\hat{k} \cdot \hat{j}$ . Since $\hat{k}$ and $\hat{j}$ are orthogonal (perpendicular) unit vectors, their dot product is zero.
$\hat{a} \cdot \hat{b} = 0 \quad \text{if } \hat{a} \perp \hat{b}$
Therefore,

$\hat{k} \cdot \hat{j} = 0$

So, the first term evaluates to $0$ .

Tip

This term is a scalar triple product $(\hat{i} \times \hat{j}) \cdot \hat{j}$ . A property of the scalar triple product is that if any two vectors are identical, the value is zero. This is because the three vectors would be coplanar, and the volume of the parallelepiped they form would be zero.

Evaluate the second term: $(\hat{j} \times \hat{i}) \cdot \hat{k}$

First, we determine the cross product $\hat{j} \times \hat{i}$ . The cross product is anti-commutative.
$\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$
Since $\hat{i} \times \hat{j} = \hat{k}$ , it follows that:

$\hat{j} \times \hat{i} = -(\hat{i} \times \hat{j}) = -\hat{k}$

Now, substitute this result back into the second term:

$(\hat{j} \times \hat{i}) \cdot \hat{k} = (-\hat{k}) \cdot \hat{k}$

Next, we evaluate the dot product $(-\hat{k}) \cdot \hat{k}$ . We can factor out the scalar constant:

$(-\hat{k}) \cdot \hat{k} = -(\hat{k} \cdot \hat{k})$

The dot product of a unit vector with itself is its magnitude squared, which is 1.
$\hat{a} \cdot \hat{a} = |\hat{a}|^2 = 1^2 = 1$
Therefore,

$\hat{k} \cdot \hat{k} = 1$

So, the second term evaluates to:

$-(\hat{k} \cdot \hat{k}) = -(1) = -1$

Watch out

A common mistake is to assume $\hat{j} \times \hat{i} = \hat{k}$ , ignoring the anti-commutativity. This would lead to $\hat{k} \cdot \hat{k} = 1$ , resulting in an incorrect final answer.

Sum the results of the two terms

The total expression is the sum of the values from Step 1 and Step 2:

$(\hat{i} \times \hat{j}) \cdot \hat{j} + (\hat{j} \times \hat{i}) \cdot \hat{k} = 0 + (-1)$

$= -1$

The final value of the expression is $-1$ .

✓Final answer
$\boxed{-1}$
CBSE 2023Set ANNUAL1 markMCQ
$\hat{i}\cdot(\hat{j}\times\hat{k})+\hat{j}\cdot(\hat{i}\times\hat{k})+\hat{k}\cdot(\hat{i}\times\hat{j})$
(a) $0$
(b) $-1$
(c) $1$
(d) $3$
›Reveal solutionSolution

Use the standard unit-vector cross products $\hat{j}\times\hat{k}=\hat{i}$ , $\hat{i}\times\hat{k}=-\hat{j}$ , $\hat{i}\times\hat{j}=\hat{k}$ .

$\hat{i}\cdot(\hat{j}\times\hat{k})=\hat{i}\cdot\hat{i}=1$

$\hat{j}\cdot(\hat{i}\times\hat{k})=\hat{j}\cdot(-\hat{j})=-1$

$\hat{k}\cdot(\hat{i}\times\hat{j})=\hat{k}\cdot\hat{k}=1$

Sum $=1-1+1=1$

✓Final answer
$1$
CBSE 2022Set ANNUAL1 mark
$[2\hat i,\ 3\hat i,\ \hat j] =$
›Reveal solutionSolution

A scalar triple product is zero whenever two of the three vectors are parallel (scalar multiples of each other).

$[2\hat i,\,3\hat i,\,\hat j]=2\hat i\cdot(3\hat i\times\hat j)$ .

$3\hat i\times\hat j=3\hat k$ (since $\hat i\times\hat j=\hat k$ ), so the product is $2\hat i\cdot3\hat k=6(\hat i\cdot\hat k)=6\cdot0=0$ .

(Equivalently: $2\hat i$ and $3\hat i$ are parallel, so the determinant/triple product is automatically $0$ .)

✓Final answer
$0$
CBSE 2022Set ANNUAL1 markMCQ
$(2\vec{i} + 3\vec{k}) \cdot (\vec{i} + \vec{j} + 4\vec{k}) \times (3\vec{i} + \vec{j} + 7\vec{k}) =$
(a) $0$
(b) $112$
(c) $126$
(d) $192$
›Reveal solutionSolution

The scalar triple product evaluates to $0$ (the three vectors are coplanar).

First, $(\hat i+\hat j+4\hat k)\times(3\hat i+\hat j+7\hat k)$ :

$\hat i:(1)(7)-(4)(1)=3$ ; $\hat j:-[(1)(7)-(4)(3)]=5$ ; $\hat k:(1)(1)-(1)(3)=-2$ .

So the cross product is $3\hat i+5\hat j-2\hat k$ .

Now dot with $(2\hat i+0\hat j+3\hat k)$ : $(2)(3)+(0)(5)+(3)(-2)=6-6=0$ .

✓Final answer
$0$
CBSE 2022Set ANNUAL1 markMCQ
$(\vec{i} + \vec{j} + \vec{k}) \cdot (\vec{i} - \vec{j} - \vec{k}) \times (\vec{i} + 2\vec{j} - \vec{k}) =$
(a) $0$
(b) $2$
(c) $4$
(d) $6$
›Reveal solutionSolution

The scalar triple product equals $6$ .

First compute $(\hat i-\hat j-\hat k)\times(\hat i+2\hat j-\hat k)$ :

$\hat i:((-1)(-1)-(-1)(2))=1+2=3$ ; $\hat j:-((1)(-1)-(-1)(1))=-(-1+1)=0$ ; $\hat k:((1)(2)-(-1)(1))=2+1=3$ .

Cross $=(3,0,3)$ . Dot with $(1,1,1)$ : $3+0+3=6$ .

✓Final answer
$6$
CBSE 2021Set I1 markMCQ
$\vec{j}\cdot(\vec{k}\times\vec{i}) =$
(a) $0$
(b) $1$
(c) $-1$
(d) $\vec{j}$
›Reveal solutionSolution

$\vec k\times\vec i=\vec j$ and $\vec j\cdot\vec j=1$ .

Using the right-handed cyclic rule, $\vec k\times\vec i=\vec j$ .

Then $\vec j\cdot(\vec k\times\vec i)=\vec j\cdot\vec j=1$ .

Equivalently, this is the scalar triple product $[\vec j\ \vec k\ \vec i]$ , a cyclic permutation of $[\vec i\ \vec j\ \vec k]=1$ .

✓Final answer
$1$

Q.The two co-initial adjacent sides of a parallelogram are 2ı̂ − 4ȷ̂ − 5k̂ and 2ı̂ + 2ȷ̂ + 3k̂ . Find its diagonals and use them to find the area of the parallelogram.

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