Question

Given Ā=-2i+ 3j+k and B=i+2j-4k. Unit vector perpendicular to the direction of A and
B:​

Answer 1

Answer :

• Unit vector perpendicular to vector $\sf{\vec{A}}$ and $\sf{\vec{B}}$ will be, $\sf{\dfrac{-2\;\hat i -2\;\hat j - \hat k}{3}}$.

Explanation :

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Given :

• $\sf{\vec{A}=-2\hat{i}+3\hat{j}+\hat{k}}$
• $\sf{\vec{B}=\hat{i}+2\hat{j}-4\hat{k}}$

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To find :

• A unit vector perpendicular to Vector $\sf{\vec{A}}$ and $\sf{\vec{B}}$

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Knowledge required :

• The cross product of two vectors gives a new vector that will be perpendicular to the two vectors.

Let, $\sf{\vec{A},\:\vec{B}\:and\:\vec{C}}$ be three vectors, as

$\sf{\vec{C}=\vec{A} \:\times \: \vec{B}}$

then, vector C will be perpendicular to vectors A and B.

• Finding Unit vector along a given vector

In general, A unit vector along the direction of given vector is be given by,

$\sf{n=\dfrac{\vec{A}}{\mid \vec{A} \mid}}$

[ where n is the unit vector along the direction of vector $\sf{\vec{A}}$ ]

• Finding the cross product of two vectors (orthogonal)

Let, two vectors be ;

$\sf{\vec{A}=a\hat{i}+b\hat{j}+c\hat{k}\;;\;\vec{B}= d\hat{i}+e\hat{j}+f\hat{k}}$ and  $\sf{\vec{C}=\vec{A} \:\times \: \vec{B}}$

then,

$\sf{\vec{C}=\vec{A}\:\times\;\vec{B}=}\left[ \begin{array} {c c c} \sf{\hat{i}} &\sf \hat{j} &\sf \hat{k} \\\sf (a) & \sf(b) & \sf(c) \\ \sf(d) &\sf (e) & \sf(f) \end{array}\right]$

$\sf{\vec{C}=\hat i(bf-ec)-\hat j(af-dc)+\hat k (ae-db)}$

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Solution :

Let, $\sf{\vec{C}}$ be the cross product of given vectors $\sf{\vec{A}}$ and $\sf{\vec{B}}$

then,

$\implies\sf{\vec{C}=\vec{A}\:\times\;\vec{B} =} \left[ \begin{array} {c c c} \sf \hat{i} & \sf\hat{j} &\sf \hat{k} \\ (-2) & (3) & (1) \\ (1) & (2) & (-4) \end{array}\right]$

$\implies\sf{\vec{C}=\hat i[ (3)(-4) -(2)(1)]-\hat j [(3)(-4)-(2)(1)]+\hat k [(-2)(2)-(1)(3)]}$

$\underline{\underline{\red{\implies\sf{\vec{C}=-14\;\hat{i} + 14\;\hat{j}-7\;\hat{k}}}}}$

Now,

Finding unit vector along the direction of $\sf{\vec{C}}$

$\implies\sf{Required \:unit\:vector=\dfrac{\vec{C}}{\mid \vec{C}\mid}}$

$\implies\sf{Required \:unit\:vector=\dfrac{-14\;\hat i +14\hat j-7\hat k}{\sqrt{(-14)^2+(14)^2+(-7)^2}}}$

$\implies\sf{Required \:unit\:vector=\dfrac{-14\;\hat i +14\hat j-7\hat k}{21}}$

$\boxed{\red{\implies\sf{Required \:unit\:vector=\dfrac{-2\;\hat i +2\hat j-\hat k}{3}}}}$

Answer

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