Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\dfrac{\pi}{3}$. Prove that $\vec{a} = \pm\dfrac{2}{\sqrt{3}}(\vec{b} \times \vec{c})$.

Let veca, vecb, vecc be unit vectors such that vec – Class 11 Mathematics

Question 89 of 113

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

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2020Subjective· 3mImportance★★★★★

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$\vec a$ is perpendicular to both $\vec b$ and $\vec c$ , so it must be a scalar multiple of $\vec b\times\vec c$ ; equating magnitudes pins down that scalar as $\pm\dfrac2{\sqrt3}$ .

Given $\vec a\cdot\vec b=0$ and $\vec a\cdot\vec c=0$ , the vector $\vec a$ is perpendicular to both $\vec b$ and $\vec c$ .

The cross product $\vec b\times\vec c$ is, by definition, also perpendicular to both $\vec b$ and $\vec c$ . In three dimensions, there is only one direction (up to sign) perpendicular to two non-parallel vectors, so $\vec a$ must be parallel to $\vec b\times\vec c$ :

$\vec a=\lambda(\vec b\times\vec c)\quad\text{for some scalar }\lambda.$

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

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

The cyclic order gives $\vec{i}\times\vec{j}=\vec{k}$ , $\vec{j}\times\vec{k}=\vec{i}$ , $\vec{k}\times\vec{i}=\vec{j}$ . Reversing the last one: …

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CBSE 2026Set ANNUAL1 markMCQ
$(2\hat{i}+3\hat{j}) \times (\hat{i}+2\hat{j})$
(a) $\hat{i}$
(b) $\hat{j}$
(c) $\hat{k}$
(d) None of these
›Reveal solutionSolution

Compute the cross product using the determinant formula; since both vectors have zero $\hat k$ -component, only the $\hat k$ term survives.

$(2\hat i+3\hat j)\times(\hat i+2\hat j) = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 0 \\ 1 & 2 & 0 \end{vmatrix}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$|\vec{a}| = 13$
(a) 45
(b) 15
(c) 25
(d) 35
›Reveal solutionSolution

With $\vec a\cdot\vec b=60$ , $|\vec a|=13$ , $|\vec b|=5$ , we get $\cos\theta=60/65=12/13$ , so $\sin\theta=5/13$ , giving $|\vec a\times\vec b|=65\times5/13=25$ .

Given $|\vec a|=13$ , $|\vec b|=5$ , and $\vec a\cdot\vec b=60$ (this is read as the dot-product value between the two vectors, since a dot product cannot equal an angle in degrees).

Since $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ : $60=13\times5\times\cos\theta=65\cos\theta \Rightarrow \cos\theta=\dfrac{60}{65}=\dfrac{12}{13}$ .

…

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

The cross product of a vector with itself is zero, so the result is $\vec{k}$ .

By the cyclic rule $\vec{i}\times\vec{j}=\vec{k}$ , and any vector crossed with itself is the zero vector, so $\vec{i}\times\vec{i}=\vec{0}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If |a⃗|=4, |b⃗|=2\sqrt3, |a⃗\timesb⃗|=12 then the angle between the vectors a⃗ and b⃗ is$
(a) π/3
(b) π/6
(c) π/4
(d) π/2
›Reveal solutionSolution

Use $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$ to solve for $\theta$ .

The magnitude of the cross product satisfies $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$ , where $\theta$ is the angle between the vectors.

$12 = (4)(2\sqrt3)\sin\theta = 8\sqrt3\,\sin\theta$

$\sin\theta = \frac{12}{8\sqrt3} = \frac{3}{2\sqrt3} = \frac{\sqrt3}{2}$

…

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CBSE 2024Set D1 markMCQ
$(\vec{i} + 3\vec{j} - 2\vec{k}) \times (-\vec{i} + 3\vec{k}) =$
(a) $9\vec{i} - \vec{j} + 3\vec{k}$
(b) $9\vec{i} + \vec{j} - 3\vec{k}$
(c) $\vec{i} - \vec{j} + 3\vec{k}$
(d) $\vec{i} + \vec{j} - 3\vec{k}$
›Reveal solutionSolution

Evaluate the determinant form of the cross product.

With $\vec{a} = (1, 3, -2)$ and $\vec{b} = (-1, 0, 3)$ :

$\vec{a}\times\vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 3 & -2 \\ -1 & 0 & 3 \end{vmatrix}.$ …

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

By the right-hand rule $\vec{k}\times\vec{j} = -\vec{i}$ ; none of the four printed options matches this, so the question as printed is defective.

Using the cyclic rule $\vec{i}\times\vec{j} = \vec{k}$ , $\vec{j}\times\vec{k} = \vec{i}$ , $\vec{k}\times\vec{i} = \vec{j}$ , and the anti-commutativity $\vec{b}\times\vec{a} = -(\vec{a}\times\vec{b})$ :

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

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CBSE 2024Set ANNUAL1 markMCQ
$\hat{k} \times \hat{j} =$
(a) $\hat{i}$
(b) $-\hat{i}$
(c) $\hat{j}$
(d) $0$
›Reveal solutionSolution

The standard cyclic products are î×ĵ=k̂, ĵ×k̂=î, k̂×î=ĵ; reversing any pair flips the sign.

…

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CBSE 2023Set E1 markMCQ
$\vec{j}\times\vec{i} =$
(a) $\vec{k}$
(b) $-\vec{k}$
(c) $\vec{0}$
(d) $1$
›Reveal solutionSolution

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

The cyclic rule gives $\vec{i}\times\vec{j}=\vec{k}$ . Reversing the order changes the sign:

…

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CBSE 2023Set E1 markMCQ
$(3\vec{k}-7\vec{i})\times 2\vec{k} =$
(a) $-14\vec{j}$
(b) $14\vec{j}$
(c) $11\vec{i}-2\vec{k}$
(d) $2\vec{k}-11\vec{i}$
›Reveal solutionSolution

Using $\vec{k}\times\vec{k}=0$ and $\vec{i}\times\vec{k}=-\vec{j}$ , the product is $14\vec{j}$ .

Distribute the cross product:

$(3\vec{k}-7\vec{i})\times 2\vec{k}=6(\vec{k}\times\vec{k})-14(\vec{i}\times\vec{k}).$

…

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CBSE 2023Set E1 markMCQ
$(10\vec{i}+\vec{j}+\vec{k})\times(-4\vec{i}+7\vec{j}-11\vec{k}) =$
(a) $-18\vec{i}+106\vec{j}+74\vec{k}$
(b) $18\vec{i}-106\vec{j}-74\vec{k}$
(c) $18\vec{i}+106\vec{j}+74\vec{k}$
(d) $5\vec{i}-6\vec{j}-7\vec{k}$
›Reveal solutionSolution

Evaluating the determinant gives $-18\vec{i}+106\vec{j}+74\vec{k}$ .

Compute

$(10\vec{i}+\vec{j}+\vec{k})\times(-4\vec{i}+7\vec{j}-11\vec{k})=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\10&1&1\\-4&7&-11\end{vmatrix}.$

$\vec{i}$ -component: $(1)(-11)-(1)(7)=-11-7=-18$ .

$\vec{j}$ -component: $-\big[(10)(-11)-(1)(-4)\big]=-[-110+4]=106$ .

…

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CBSE 2023Set ANNUAL1 markMCQ
$\hat{i} \times \hat{j} =$
(a) 0
(b) $\hat{k}$
(c) $-\hat{k}$
(d) none of these
›Reveal solutionSolution

This is one of the fundamental unit-vector cross-product identities.

…

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Q.Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\dfrac{\pi}{3}$ . Prove that $\vec{a} = \pm\dfrac{2}{\sqrt{3}}(\vec{b} \times \vec{c})$ .

Concept understanding — Vector (Cross) Product

Vector (Cross) Product

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