Scalar or Dot Product MCQ Quiz - Objective Question with Answer for Scalar or Dot Product - Download Free PDF

Calculation:

Given,

\( \cos^2(\alpha) + \cos^2(\beta) + \cos^2(\gamma) = 1 \)

Using the identity \( \cos^2(x) = 1 - \sin^2(x) \), we substitute:

\( (1 - \sin^2(\alpha)) + (1 - \sin^2(\beta)) + (1 - \sin^2(\gamma)) = 1 \)

Simplifying the equation:

\( 3 - (\sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma)) = 1 \)

Rearrange to isolate the sine terms:

\( \sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) = 2 \)

Now, calculate the dot product:

\( \vec{a} \cdot \vec{b} = \sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) = 2 \)

∴ The value of \( \vec{a} \cdot \vec{b} \)is 2.

Hence, the correct answer is Option 4.

Calculation: 

\(\overrightarrow{\mathrm{b}}=\lambda \overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{a}} \times \hat{\mathrm{j}})\)

⇒ \(\overrightarrow{\mathrm{b}}=\lambda(-2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})\)

⇒ \(|\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}| \quad \therefore \sqrt{6}=\sqrt{12}|\lambda| \Rightarrow \lambda= \pm \frac{1}{\sqrt{2}}\)

\(\left(\lambda=\frac{1}{\sqrt{2}} \text { rejected } \because \overrightarrow{\mathrm{b}} \text { makes acute angle with y axis }\right)\)

⇒ \(\overrightarrow{\mathrm{b}}=-\sqrt{2}(-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\)

⇒ \(\frac{(3 \vec{a}+\sqrt{2 b}) \cdot \vec{c}}{|\vec{c}|}=3 \sqrt{2}\)

Hence, the correct answer is Option 1.

Explanation:

Given:

\(⃗ a+2⃗ b\) and \(\rm 5⃗ a-4⃗ b\) are perpendicular vectors.

⇒ \((⃗ a+2⃗ b).\)\((\rm 5⃗ a-4⃗ b)\) = 0

⇒ \(5|⃗ a|^2 -4⃗ a.⃗ b + 10⃗ a . ⃗ b - 8(⃗ b)^2=0\)

⇒ \(5× 1 + 6⃗ a.⃗ b-8 =0\)

Now \(\vec a, \vec b\) are unit vectors

⇒ \(6 |\vec a||\vec b| Cosθ = 3\)

⇒ 1.1 cosθ =1/2

⇒Cosθ = 1/2 

Now

cos2 θ = Cos²θ -1

= 2× 1/4 -1

= \(-\frac{1}{2}\)

Now, 

cosθ + cos2θ = 1/2 -1/2 =0

∴The Correct answer is Option a

Concept:

\(\begin{aligned} |\vec{a}-\vec{b}|^{2} &=|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b} \\ &=|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}-4 \vec{a} \cdot \bar{b} \\ &=|\vec{a}+\vec{b}|^{2}-4 \vec{a} \cdot \vec{b} \end{aligned}\)​

\({\rm{\vec a}}.{\rm{\vec b}}= |{\rm{\vec a}}|.|{\rm{\vec b}}|\cos \theta\)

Vector a is perpendicular to b if \({\rm{\vec a}}.{\rm{\vec b}}= 0\)

Calculation:

\(\left| {{\rm{\vec a}} + {\rm{\vec b}}} \right| = \left| {{\rm{\vec a}} - {\rm{\vec b}}} \right|\)

Squaring both sides,

\(\begin{aligned} |\vec{a}-\vec{b}|^{2} &=|\vec{a}+\vec{b}|^{2}\end{aligned}\)

Now we have,

\(\begin{aligned} |\vec{a}-\vec{b}|^{2} &=|\vec{a}+\vec{b}|^{2}-4 \vec{a} \cdot \vec{b} =|\vec{a}+\vec{b}|^{2}\end{aligned}\\\Rightarrow-4 \vec{a} \cdot \vec{b}=0 \\\Rightarrow \vec{a} \cdot \vec{b} =0\)

∴  Vector a is perpendicular to b

Hence, option (3) is correct.

Calculation

Given: \(\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}\) and \(\mathbf{b} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\) 

\((\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b}\)

⇒ \((\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b} = |\mathbf{a}|^2 - |\mathbf{b}|^2\)

\(|\mathbf{a}|^2 = (1)^2 + (1)^2 + (1)^2 = 1 + 1 + 1 = 3\)

\(|\mathbf{b}|^2 = (1)^2 + (2)^2 + (3)^2 = 1 + 4 + 9 = 14\)

\(|\mathbf{a}|^2 - |\mathbf{b}|^2 = 3 - 14 = -11\)

⇒ \((\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = -11\).

Hence option 2 is correct.

Concept:

Dot Product: it is also called the inner product or scalar product

Let the two vectors are \(\rm \vec a\) and \(\rm \vec b\)

Dot Product of two vectors is given by:  \(\rm \vec a.{\rm{\;}}\vec b\) = |a||b| cos θ

Where |\(\rm \vec a\)| = Magnitudes of vectors a, |\(\rm \vec b\)| = Magnitudes of vectors b and θ is the angle between a and b

Formulas of Dot Product:

 \(\rm \vec i.\vec i = \vec j.\vec j = \vec k.\vec k = 1\)

\(\rm \vec i.\vec j = \vec j.\vec i = \vec i.\vec k = \vec k.\vec i =\vec j.\vec k= \vec k.\vec j = 0\)

Calculation:

Given that,

(â + b̂ + ĉ) = 0    ----(1)

We know that,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

⇒ (â + b̂ + ĉ)² = â ⋅ â + b̂ ⋅ b̂ + ĉ ⋅ ĉ + 2(â ⋅ b̂ + b̂ ⋅ ĉ + ĉ ⋅ â) 

From equation (1), we get

⇒ (1 + 1 + 1) + 2 (â ⋅ b̂ + b̂ ⋅ ĉ + ĉ ⋅ â) = 0

∴ (â ⋅ b̂ + b̂ ⋅ ĉ + ĉ ⋅ â) = - 3/2        

Concept:

If  \(\rm \vec{a}\) and \(\rm \vec{b}\) are mutually perpendicular unit vectors \(\rm \vec{a}.\vec{b} = 0= ​​\)\(\rm \vec{b}.\vec{a} = 0\)

\(\rm \vec{a}.\vec{a} = |\vec{a}|^2\),  \(\rm \vec{b}.\vec{b} = |\vec{b}|^2\)

Calculations:

Consider, \(\rm (3\vec{a}+2\vec{b})\cdot (5\vec{a}-6\vec{b})\)

Given  \(\rm \vec{a}\) and \(\rm \vec{b}\) are mutually perpendicular unit vectors.

So,  \(\rm \vec{a}.\vec{b} = 0= ​​\)\(\rm \vec{b}.\vec{a} = 0\)

And   \(\rm \vec{a}\) and \(\rm \vec{b}\) are unit vectors

So, \(\rm |\vec{a}|= |\vec{b}| = 1\)

\(\rm (3\vec{a}+2\vec{b})\cdot (5\vec{a}-6\vec{b})\)

\(= \rm 3\vec{a}\cdot 5\vec{a}+2\vec{b}\cdot 5\vec{a} -3\vec{a}\cdot6\vec{b}-2\vec{b}\cdot 6\vec{b}\)

= \(\rm 15|\vec {a}|^2 - 12 |\vec {b}|^2\)

= 15 - 12 

= 3

Concept:

Dot Product of two vectors \(\rm \vec A\) and \(\rm \vec B\) is defined as \(\rm \vec A.\vec B=\rm \left|\vec A\right|\rm \left|\vec B\right|\cos \theta\), where \(\rm \left|\vec A\right|\) is the magnitude of vector \(\rm \vec A\).

\(\rm \vec A.\vec A=\rm \left|\vec A\right|^2\).

Calculation:

We are given that "sum of two vectors \(\rm \vec a\) and \(\rm \vec b\) is a vector \(\rm \vec c\)".

⇒ \(\rm \vec a +\vec b=\vec c\)

Taking dot product of both sides with themselves, the magnitudes will still be equal:

⇒ \(\rm \left(\vec a +\vec b\right).\left(\vec a +\vec b\right)=\left(\vec c\right).\left(\vec c\right)\)

⇒ \(\rm \left|\vec a\right|^2+\left|\vec b\right|^2+2\vec a.\vec b=\left|\vec c\right|^2\)

Since \(\rm \left|\vec a \right|=\left|\vec b \right|=\left|\vec c \right|=2\), we get:

⇒ \(\rm 2^2+2^2+2\vec a.\vec b=2^2\)

⇒ \(\rm 4+4+2\vec a.\vec b=4\)

⇒ \(\rm 2\vec a.\vec b=-4\)

Now, \(\rm \rm \left|\vec a-\vec b\right|^2=\left(\vec a -\vec b\right).\left(\vec a -\vec b\right)\)

= \(\rm \left|\vec a\right|^2+\left|\vec b\right|^2-2\vec a.\vec b\)

= 4 + 4 - (-4)

= 12

⇒ \(\rm \rm \left|\vec a-\vec b\right|=\sqrt{12}=2\sqrt3\).

Concept:

If \(\rm \vec{a} \ and \ \vec b\) are two vectors then \(\rm \vec{a}.\vec{b}=|\vec{a}||\vec{b}|cosθ\)

Note: If vectors \(\rm \vec{a} \ and \ \vec b\) are perpendicular to each other then \(\rm \vec{a}.\vec{b}=0\)

Calculation:

Given: \(\vec a = \hat i + \hat j - \hat k \ and \ \vec b =\hat i - \hat j - \hat k\)

Let θ be the angle between the vector \(\rm \vec{a} \ and \ \vec b\)

⇒ \(|\vec a| = \sqrt 3 \ and \ |\vec b| = \sqrt 3\)

We know that, 

\(\rm \vec{a}.\vec{b}=|\vec{a}||\vec{b}|cosθ\)

⇒ \((\hat i + \hat j - \hat k) \cdot (\hat i - \hat j - \hat k) = \sqrt 3 \times \sqrt 3 \times cos θ \)

⇒ 1 = 3 cos θ 

⇒ \(\rm cos\ θ=\frac{1}{3}\) 

⇒ \(\rm θ=cos^{-1}\frac{1}{3}\)

Hence, option 3 is correct.

CONCEPT:

• \(\vec a \cdot \vec a = |\vec a|^2\)
• \(\vec a \cdot \vec b = \vec b \cdot \vec a\)
• If \(\vec a\) is a unit vector then \(|\vec a| = 1\)

CALCULATION:

Given: \((\vec x - \vec a) \cdot (\vec x + \vec a) = 12\) and \(\vec a\) is a unit vector

⇒ \((\vec x - \vec a) \cdot (\vec x + \vec a) = |\vec x|^2 + \vec x \cdot \vec a - \vec a \cdot \vec x - |\vec a|^2 = 12\)

As we know that, \(\vec a \cdot \vec b = \vec b \cdot \vec a\)

⇒ \((\vec x - \vec a) \cdot (\vec x + \vec a) = |\vec x|^2 - |\vec a|^2 = 12\)

As we know that, if \(\vec a\) is a unit vector then \(|\vec a| = 1\)

⇒ \(|\vec x|^2 = 13 \Rightarrow |\vec x| = \sqrt {13}\)

Hence, correct option is 2.

Concept:

• The cross product of vector to itself = 0
• The cross product of collinear vectors = 0
• The dot product of collinear vectors = Product of their Magnitudes
• \(\rm \vec a \times \vec b=-\vec b \times \vec a\)
• For dot product \(\rm (\vec P + \vec Q) \cdot \vec R = (\vec P\cdot \vec R) +(\vec Q\cdot \vec R)\)
• For cross product \(\rm (\vec P + \vec Q) ×\vec R = (\vec P×\vec R) + (\vec Q×\vec R) \) 
• The unit vector in the direction of a \(\rm \vec P= \hat P={\vec P\over\left|\vec P\right|}\) 
• A vector \(\rm \vec X\) in direction of \(\rm \vec P\) = (Magnitude of \(\rm \vec X\)) × \(\rm \hat P\)

Calculation:

Given:

\(\rm\vec a × (2\hat i + 3\hat j + 4\hat k) = (2\hat i + 3\hat j + 4\hat k) × \vec b\)

⇒ \(\rm\vec a × (2\hat i + 3\hat j + 4\hat k) - (2\hat i + 3\hat j + 4\hat k) × \vec b=0\)

⇒ \(\rm\vec a × (2\hat i + 3\hat j + 4\hat k) + \vec b × (2\hat i + 3\hat j + 4\hat k) =0\)

⇒ \(\boldsymbol{\rm(\vec a + \vec b) × (2\hat i + 3\hat j + 4\hat k) =0}\)

It means the \(\rm\vec a + \vec b\) and \(\rm2\hat i + 3\hat j + 4\hat k\) are collinear vectors.

∴ \(\rm\vec a + \vec b = \left|\vec a + \vec b\right| × {2\hat i + 3\hat j + 4\hat k\over \left|2\hat i + 3\hat j + 4\hat k\right|}\) 

⇒ \(\rm\vec a + \vec b = \sqrt{29}× {2\hat i + 3\hat j + 4\hat k\over\sqrt{29}}\) 

⇒ \(\boldsymbol{\rm\vec a + \vec b = 2\hat i + 3\hat j + 4\hat k}\)

\(\rm (\vec a + \vec b)\cdot(-7\hat i + 2\hat j + 3\hat k) = \left(2\hat i + 3\hat j + 4\hat k\right)\cdot\left(-7\hat i + 2\hat j + 3\hat k\right)\)

⇒ \(\rm (\vec a + \vec b)\cdot(-7\hat i + 2\hat j + 3\hat k) = \left(2\times(-7) + 3\times2 + 4\times3\right)\) 

⇒ \(\rm (\vec a + \vec b)\cdot(-7\hat i + 2\hat j + 3\hat k) = \left(-14 +6 + 12\right)\) 

⇒ \(\boldsymbol{\rm (\vec a + \vec b)\cdot(-7\hat i + 2\hat j + 3\hat k) = 4}\)

Concept:

Dot Product: it is also called the inner product or scalar product

Let the two vectors are \(\rm \vec a\) and \(\rm \vec b\)

Dot Product of two vectors is given by:  \(\rm \vec a.{\rm{\;}}\vec b\) = |a||b| cos θ

Where |\(\rm \vec a\)| = Magnitudes of vectors a, |\(\rm \vec b\)| = Magnitudes of vectors b and θ is the angle between a and b

Formulas of Dot Product:

 \(\rm \vec i.\vec i = \vec j.\vec j = \vec k.\vec k = 1\)

\(\rm \vec i.\vec j = \vec j.\vec i = \vec i.\vec k = \vec k.\vec i =\vec j.\vec k= \vec k.\vec j = 0\)

Calculation:

Let \(\vec{a}\ =\ x̂{i}\ +\ ŷ{j}\ +\ ẑ{k}\)

\(\Rightarrow\ |a|\ =\ \sqrt{x^2\ +\ y^2\ +\ z^2}\)

\(\vec{a}.̂{i}\ =\ (x̂{i}\ +\ ŷ{j}\ +\ ẑ{k}).̂{i}\ =\ x\)

Similarly, \(\vec{a}.̂{j}\ =\ y, \ \vec{a}.̂{k}\ =\ z\)

Therefore \([(\vec{a}.̂{i})̂{i}\ +\ (\vec{a}.̂{j})̂{j}\ +\ (\vec{a}.̂{k})̂{k}]\)

= xî + yĵ + zk̂ 

Hence, required value of

 \([(\vec{a}.\hat{i})\hat{i}\ +\ (\vec{a}.\hat{j})\hat{j}\ +\ (\vec{a}.\hat{k})\hat{k}].\vec{a}\)

= (xî + yĵ + zk̂ ).(xî + yĵ + zk̂ )

= x² + y² + z² = |a|²

CONCEPT:

• Projection of a vector \(\vec a\) on other vector \(\vec b\) is given by: \(\vec a \cdot \hat b = \frac{\vec a \cdot \vec b}{|\vec b|}\)

CALCULATION:

Given: \(\vec a = 2\hat i + 3\hat j + 2\hat k\) and \(\vec b = \vec i + 2\vec j + \hat k\)

Here, we have to find the projection of a vector \(\vec a\) on other vector \(\vec b\) is given by: \(\vec a \cdot \hat b = \frac{\vec a \cdot \vec b}{|\vec b|}\)

⇒ \(\vec a \cdot \vec b = 2 + 6 + 2 = 10 \ and \ |\vec b| = \sqrt {6}\)

⇒ \(\vec a \cdot \hat b = \frac{10}{\sqrt {6}} = \frac{5\sqrt6}{3}\)

Hence, option 3 is correct.

CONCEPT:

The scalar product of two vectors \(\vec a \ and \ \vec b \)is given by \(\vec a \cdot \;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right|\cos θ \)

If the vectors \(\vec a \ and \ \vec b \) are perpendicular then \(\vec a \cdot \;\vec b = 0\)

CALCULATION:

Given: \(\vec a \:and\: \vec b\) are two vectors such that \(|\vec a + \vec b|= |\vec a - \vec b|=4\)

⇒ \(|\vec a + \vec b|^2= |\vec a - \vec b|^2\)

⇒ \((\vec a + \vec b) \cdot (\vec a + \vec b) = (\vec a - \vec b) \cdot (\vec a - \vec b)\)

⇒ \(|\vec a|^2 + \vec a \cdot \vec b + \vec b \cdot \vec a + |\vec b|^2 = |\vec a|^2 - \vec a \cdot \vec b - \vec b \cdot \vec a + |\vec b|^2\)

∵ \(\vec a \cdot \vec b = \vec b \cdot \vec a\)

⇒ \(|\vec a|^2 + 2\vec a \cdot \vec b + |\vec b|^2 = |\vec a|^2 - 2\vec a \cdot \vec b + |\vec b|^2\)

⇒ \(4\vec a \cdot \vec b = 0\)

⇒ \(\vec a \cdot \vec b = 0\)

So, \(\vec a \) must be perpendicular to \(\vec b.\)

Hence, correct option is 3.

Concept:

If \(\rm(\vec {a} + \vec{b})\) is perpendicular to \(\rm\vec {a}\), then 

\(\rm(\vec {a} + \vec{b})\).\(\vec {a}\) = 0

Calculation:

\(|\vec{a}|^{2}\) + \(\rm \vec {a}.\rm \vec {b}\) = 0

Given: \(\rm |\vec{b}| = 2\times |\vec{a}|\)

So, we can write

\(\rm \frac{|\vec{b}|^{2}}{4}\) +  \(\rm \vec {a}.\rm \vec {b}\) = 0

\(\rm \vec {a}.\rm \vec {b}\) = - \(\rm \frac{|\vec{b}|^{2}}{4}\)      -----(1)

To find: \(\rm (4\vec {a} + \vec{b})\cdot \vec{b}\)

\(\rm (4\vec {a} + \vec{b})\cdot \vec{b}\) = 4\(\rm \vec {a}.\rm \vec {b}\) + \(\rm |\vec{b}|^{2}\)      ----(2)

From equations (1) & (2) we can write,

​\(\rm (4\vec {a} + \vec{b})\cdot \vec{b}\) = -​\(\rm |\vec{b}|^{2}\) + \(\rm |\vec{b}|^{2}\) = 0​​