The angle θ between the vectors a → = 5 i ^ − j ^ + k ^ and b → = i ^ + j ^ − k ^ is:

Explanation: We have two vectors a → = 5 i ^ − j ^ + k ^ and b → = i ^ + j ^ − k ^ We know that the angle θ between two vectors a → and a → is: θ = cos − 1 ⁡ a → ⋅ b → | a → | × | b | ----(1)Therefore, a → ⋅ b → = ( 5 i ^ − j ^ + k ^ ) ⋅ ( i ^ + j ^ − k ^ ) ⇒ a → ⋅ b → = 5 i ^ 2 + 5 i ^ j ^ − 5 i ^ k ^ − i ^ j ^ − j ^ 2 + j ^ k ^ + i ^ k ^ + j ^ k ^ − k ^ 2 We know that, ( i ^ 2 = j ^ 2 = k ^ 2 = 1 ) and ( i ^ j ^ = j ^ k ^ = i ^ k ^ = 0 ) ∴ a → ⋅ b → = 5 − 1 − 1 = 3 We also know that, if a → = ( x i ^ + y j ^ + z k ^ ) then, | a → | = x 2 + y 2 + z 2 Therefore, according to the question, ⇒ | a → | = 5 2 + ( − 1 ) 2 + 1 2 = 27 ⇒ | b → | = 1 2 + 1 2 + ( − 1 ) 2 = 3 Putting all the above values in equation (1), ⇒ θ = cos − 1 ⁡ 3 27 × 3 ⇒ θ = cos − 1 ⁡ 1 3 Hence, the correct option is (A).

Question:

The angle $θ$ between the vectors $\vec{a} = 5 \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$ is:

JEE Main

Updated On: Aug 21, 2024

(A) $\cos^{- 1} (\frac{1}{3})$
(B) $\cos^{- 1} (\frac{2}{3})$
(C) $\cos^{- 1} (\frac{4}{7})$
(D) None of these

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The Correct Option is A

Approach Solution - 1

Explanation:
We have two vectors

\vec{a} = 5 \hat{i} - \hat{j} + \hat{k}

and

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

We know that the angle

θ

between two vectors

\vec{a}

and

\vec{a}

is:

θ = \cos^{- 1} \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | \times | b |}

----(1)Therefore,

\vec{a} \cdot \vec{b} = (5 \hat{i} - \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k})

\Rightarrow \vec{a} \cdot \vec{b} = 5 {\hat{i}}^{2} + 5 \hat{i} \hat{j} - 5 \hat{i} \hat{k} - \hat{i} \hat{j} - {\hat{j}}^{2} + \hat{j} \hat{k} + \hat{i} \hat{k} + \hat{j} \hat{k} - {\hat{k}}^{2}

We know that,

({\hat{i}}^{2} = {\hat{j}}^{2} = {\hat{k}}^{2} = 1)

and

(\hat{i} \hat{j} = \hat{j} \hat{k} = \hat{i} \hat{k} = 0)

∴ \vec{a} \cdot \vec{b} = 5 - 1 - 1 = 3

We also know that, if

\vec{a} = (x \hat{i} + y \hat{j} + z \hat{k})

then,

| \vec{a} | = \sqrt{x^{2} + y^{2} + z^{2}}

Therefore, according to the question,

\Rightarrow | \vec{a} | = \sqrt{5^{2} + (- 1)^{2} + 1^{2}} = \sqrt{27}

\Rightarrow | \vec{b} | = \sqrt{1^{2} + 1^{2} + (- 1)^{2}} = \sqrt{3}

Putting all the above values in equation (1),

\Rightarrow θ = \cos^{- 1} \frac{3}{\sqrt{27} \times \sqrt{3}}

\Rightarrow θ = \cos^{- 1} \frac{1}{3}

Hence, the correct option is (A).

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Approach Solution -2

Vectors are the physical quantities that have both magnitude and direction and also they obey triangle law or parallelogram laws of vector addition.

There are two ways of multiplying vectors

Scalar product
Vector product

Scalar Product

The scalar product or Dot product of any two vectors A and B denoted as A.B and given by the product of the magnitude of both the vectors multiplied by the cosine of the angle between them.

A.B=AB cos

Vector Product

The vector product or Cross product of any two vectors A and B denoted as AB and given by the product of the magnitude of both the vectors multiplied by the sine of the angle between them.

AB=AB sin

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