Find a vector of magnitude 6, which is perpendicular to both the vectors 2hati-hatj+2hatk and 4hati-hatj+3hatk

NCERT Exemplar · Q8

2\hat{i}-\hat{j}+2\hat{k}

Assam AhsecShort· 3mImportance★★★★★

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Concept understanding — Cross Product Normalization

Cross Product Normalization (Unit Vector Perpendicular to Two Vectors)

A frequent job in vector algebra is: given two vectors, find a vector of length $1$ that is perpendicular to both of them. The cross product does the perpendicular part; normalization does the length part. Putting them together is what this idea is about.

Step 1 — the cross product gives the direction

For two non-parallel vectors $\vec{a}$ and $\vec{b}$ , the cross product

\vec{a}\times\vec{b} = |\vec{a}|\,|\vec{b}|\sin\theta\;\hat{n}

is a vector that is perpendicular to both $\vec{a}$ and $\vec{b}$ . Its direction is fixed by the right-hand rule. So $\vec{a}\times\vec{b}$ already points the way we want — but its length is $|\vec{a}||\vec{b}|\sin\theta$ , which is usually not $1$ .

Step 2 — normalize to get unit length

To normalize any non-zero vector means to divide it by its own magnitude, producing a vector of length $1$ in the same direction. Applying that to the cross product:

\hat{n} = \pm\,\frac{\vec{a}\times\vec{b}}{\,|\vec{a}\times\vec{b}|\,}

This $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ . The $\pm$ matters: there are exactly two such unit normals, pointing in opposite directions. The plus sign gives the right-hand-rule direction of $\vec{a}\times\vec{b}$ ; the minus sign gives the other side.

Why the division works

Dividing by the magnitude only rescales the vector — it never changes its direction. So $\hat n$ keeps the perpendicularity that the cross product built in, while its length becomes $\dfrac{|\vec a\times\vec b|}{|\vec a\times\vec b|}=1$ .

Quick example

Let $\vec{a}=\hat{i}+\hat{j}$ and $\vec{b}=\hat{j}+\hat{k}$ . Then

\vec{a}\times\vec{b} = \hat{i}-\hat{j}+\hat{k},\qquad |\vec{a}\times\vec{b}| = \sqrt{1+1+1}=\sqrt{3}.

So a unit vector perpendicular to both is …

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$\vec{u}\times\vec{v}=-\hat{i}+2\hat{j}+2\hat{k}$ has magnitude $3$ ; scaling to length $6$ gives $\pm(-2\hat{i}+4\hat{j}+4\hat{k})$ .

The idea

The cross product of two vectors is perpendicular to both of them. So to find a vector perpendicular to both given vectors, compute their cross product to fix the direction, then rescale that direction to the required magnitude $6$ .

Step 1: cross product

Let $\vec{u}=2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{v}=4\hat{i}-\hat{j}+3\hat{k}$ .

\vec{u}\times\vec{v}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-1&2\\4&-1&3\end{vmatrix}

=\hat{i}\big[(-1)(3)-(2)(-1)\big]-\hat{j}\big[(2)(3)-(2)(4)\big]+\hat{k}\big[(2)(-1)-(-1)(4)\big]

=\hat{i}(-3+2)-\hat{j}(6-8)+\hat{k}(-2+4)=-\hat{i}+2\hat{j}+2\hat{k}

Step 2: its magnitude

|\vec{u}\times\vec{v}|=\sqrt{(-1)^2+2^2+2^2}=\sqrt{9}=3

Step 3: scale to magnitude 6 …

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

AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL4 marks
$\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}$
›Reveal solutionSolution

Since $\vec d$ is perpendicular to both $\vec a$ and $\vec b$ , it is parallel to $\vec a\times\vec b$ ; scale using $\vec c\cdot\vec d=15$ .

$\vec a=\hat i+4\hat j+2\hat k$ , $\vec b=3\hat i-2\hat j+7\hat k$ . Compute $\vec a\times\vec b$ :

$\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\1&4&2\\3&-2&7\end{vmatrix}=\hat i(4\cdot7-2\cdot(-2))-\hat j(1\cdot7-2\cdot3)+\hat k(1\cdot(-2)-4\cdot3)$

$=\hat i(28+4)-\hat j(7-6)+\hat k(-2-12)=32\hat i-\hat j-14\hat k.$

Since $\vec d\perp\vec a$ and $\vec d\perp\vec b$ , $\vec d=t(32\hat i-\hat j-14\hat k)$ for some scalar $t$ .

Now use $\vec c\cdot\vec d=15$ with $\vec c=2\hat i-\hat j+4\hat k$ : …

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AHSEC Higher Secondary (HS) Final Examination 2023Set ANNUAL4 marks
$\vec{a}+\vec{b}$
(ii) Evaluate the product $(3\vec{a}-5\vec{b}) \cdot (2\vec{a}+7\vec{b})$ . OR Show that the points $A(1, -2, -8)$ , $B(5, 0, -2)$ and $C(11, 3, 7)$ are collinear and find the ratio in which $B$ divides $AC$ .
›Reveal solutionSolution

(i) Cross product of $\vec a+\vec b$ and $\vec a-\vec b$ , then normalize; (ii) expand the dot product using known magnitudes/dot product of $\vec a,\vec b$ .

Given $\vec a = 3\hat i+2\hat j+2\hat k$ , $\vec b = \hat i+2\hat j-2\hat k$ .

(i) $\vec a+\vec b = 4\hat i+4\hat j+0\hat k$ , $\vec a-\vec b = 2\hat i+0\hat j+4\hat k$ .

$(\vec a+\vec b)\times(\vec a-\vec b) = \begin{vmatrix}\hat i&\hat j&\hat k\\4&4&0\\2&0&4\end{vmatrix} = \hat i(4\cdot4-0\cdot0) - \hat j(4\cdot4-0\cdot2) + \hat k(4\cdot0-4\cdot2) = 16\hat i-16\hat j-8\hat k.$

Magnitude $=\sqrt{16^2+16^2+8^2}=\sqrt{256+256+64}=\sqrt{576}=24$ .

Unit vector $=\dfrac{16\hat i-16\hat j-8\hat k}{24} = \dfrac{2\hat i-2\hat j-\hat k}{3}$ .

(ii) First find $|\vec a|^2 = 9+4+4=17$ , $|\vec b|^2=1+4+4=9$ , $\vec a\cdot\vec b = 3(1)+2(2)+2(-2)=3+4-4=3$ .

$(3\vec a-5\vec b)\cdot(2\vec a+7\vec b) = 6\vec a\cdot\vec a + 21\vec a\cdot\vec b - 10\vec b\cdot\vec a - 35\vec b\cdot\vec b = 6|\vec a|^2 + 11(\vec a\cdot\vec b) - 35|\vec b|^2$

$= 6(17) + 11(3) - 35(9) = 102+33-315 = -180.$

OR: $A(1,-2,-8)$ , $B(5,0,-2)$ , $C(11,3,7)$ .

…

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AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL4 marks
$\vec{a}+\vec{b}$
›Reveal solutionSolution

A unit vector perpendicular to two vectors is their normalized cross product.

$\vec a=3\hat i+2\hat j+2\hat k$ , $\vec b=\hat i+2\hat j-2\hat k$ .

$\vec a+\vec b = 4\hat i+4\hat j+0\hat k$ , $\quad \vec a-\vec b = 2\hat i+0\hat j+4\hat k$

$(\vec a+\vec b)\times(\vec a-\vec b) = \begin{vmatrix}\hat i&\hat j&\hat k\\4&4&0\\2&0&4\end{vmatrix}$

$=\hat i(4\cdot4-0\cdot0)-\hat j(4\cdot4-0\cdot2)+\hat k(4\cdot0-4\cdot2)$

$=16\hat i-16\hat j-8\hat k$

Magnitude $=\sqrt{16^2+16^2+8^2}=\sqrt{256+256+64}=\sqrt{576}=24$

Unit vector $=\dfrac{16\hat i-16\hat j-8\hat k}{24}=\dfrac13(2\hat i-2\hat j-\hat k)$ (or its negative).

OR: $\vec a=\hat i+4\hat j+2\hat k$ , $\vec b=3\hat i-2\hat j+7\hat k$ , $\vec c=2\hat i-\hat j+4\hat k$ . Find $\vec d$ perpendicular to both $\vec a,\vec b$ with $\vec c\cdot\vec d=15$ .

Since $\vec d\perp\vec a$ and $\vec d\perp\vec b$ , $\vec d$ is parallel to $\vec a\times\vec b$ , so $\vec d=\lambda(\vec a\times\vec b)$ .

…

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Q.Find a vector of magnitude 6, which is perpendicular to both the vectors $2\hat{i}-\hat{j}+2\hat{k}$ and $4\hat{i}-\hat{j}+3\hat{k}$ .

Concept understanding — Cross Product Normalization

Cross Product Normalization (Unit Vector Perpendicular to Two Vectors)

Step 1 — the cross product gives the direction

Step 2 — normalize to get unit length

Why the division works

Quick example

The idea

Step 1: cross product

Step 2: its magnitude

Step 3: scale to magnitude 6 …

Method: Vector of given length perpendicular to two vectors

Steps

Common Mistakes

Q.Find a vector of magnitude 6, which is perpendicular to both the vectors 2i^−j^+2k^2\hat{i}-\hat{j}+2\hat{k}2i^−j^​+2k^ and 4i^−j^+3k^4\hat{i}-\hat{j}+3\hat{k}4i^−j^​+3k^.

The idea

Step 1: cross product

Step 2: its magnitude

Step 3: scale to magnitude 6 …

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Q.Find a vector of magnitude 6, which is perpendicular to both the vectors $2\hat{i}-\hat{j}+2\hat{k}$ and $4\hat{i}-\hat{j}+3\hat{k}$ .

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