Let the angle $\theta, 0 < \theta < \frac{\pi}{2}$ between two unit vectors $\hat{a}$ and $\hat{b}$ be $\sin^{-1}\left(\frac{\sqrt{65}}{9}\right)$. If the vector $\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$, then the value of $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$ is Options: (A) $31$ (B) $27$ (C) $29$ (D) $24$

### Analyzing the Foundation We are given two unit vectors, $\hat{a}$ and $\hat{b}$, with an angle $\theta$ between them such that $\sin \theta = \frac{\sqrt{65}}{9}$. Since $\hat{a}$ and $\hat{b}$ are unit vectors, their magnitudes are $|\hat{a}| = 1$ and $|\hat{b}| = 1$. To find the dot product $\hat{a} \cdot \hat{b} = \cos \theta$, we use the trigonometric identity $\cos^2 \theta = 1 - \sin^2 \theta$. Substituting the given value: $$\cos^2 \theta = 1 - \frac{65}{81} = \frac{16}{81}$$ Assuming $\theta$ is acute, we take the positive root to obtain: $$\cos \theta = \frac{4}{9}$$ ### The Deconstruction of $\vec{c}$ The vector $\vec{c}$ is defined as: $$\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$$ When calculating the dot product $\vec{c} \cdot \hat{a}$, we distribute the operation across the sum: $$\vec{c} \cdot \hat{a} = 3(\hat{a} \cdot \hat{a}) + 6(\hat{b} \cdot \hat{a}) + 9((\hat{a} \times \hat{b}) \cdot \hat{a})$$ Because $\hat{a} \cdot \hat{a} = 1$ and the scalar triple product $(\hat{a} \times \hat{b}) \cdot \hat{a} = 0$ (due to the orthogonality of the cross product), the expression simplifies to: $$\vec{c} \cdot \hat{a} = 3(1) + 6\left(\frac{4}{9}\right) = 3 + \frac{8}{3} = \frac{17}{3}$$ ### Calculating the Second Component Next, we evaluate $\vec{c} \cdot \hat{b}$ using the same distributive property: $$\vec{c} \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6(\hat{b} \cdot \hat{b}) + 9((\hat{a} \times \hat{b}) \cdot \hat{b})$$ Again, the term $(\hat{a} \times \hat{b}) \cdot \hat{b}$ vanishes because the cross product is perpendicular to $\hat{b}$. We calculate the remaining terms: $$\vec{c} \cdot \hat{b} = 3\left(\frac{4}{9}\right) + 6(1) = \frac{4}{3} + 6 = \frac{22}{3}$$ ### Final Calculation We now substitute these results into the target expression $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$: $$9\left(\frac{17}{3}\right) - 3\left(\frac{22}{3}\right)$$ Simplifying the arithmetic: $$3(17) - 22 = 51 - 22 = 29$$ The final value of the expression is **29**.

Can I just find the magnitude of $\vec{c}$ first?

You could, but that would be a trap! Calculating the magnitude of $\vec{c}$ involves squaring the expression, which leads to messy terms like $(\hat{a} \times \hat{b}) \cdot \hat{a}$, which is zero, but also $(\hat{a} \times \hat{b}) \cdot (\hat{a} \times \hat{b})$. It is much faster to distribute the dot product directly into the linear expression.

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Animated Solution for Mathematics - Vector Algebra: Let the angle $θ, 0 < θ < \frac{π}{2}$ between two unit vectors $\overset{a}{^}$ and $\hat{b}$ be $sin^{- 1} (\frac{65}{9})$ . If the vector $c = 3 \overset{a}{^} + 6 \hat{b} + 9 (\overset{a}{^} \times \hat{b})$ , then the value of $9 (c \cdot \overset{a}{^}) - 3 (c \cdot \hat{b})$ is

Q: Wait, why do we reject the negative root for $\cos \theta$?

Great catch! The problem explicitly states $0 < \theta < \frac{\pi}{2}$. In the first quadrant, the cosine function is strictly positive. If we had chosen the negative root, we would be describing an obtuse angle, which contradicts the given constraints.

Q: Why is the dot product of the cross product with its own components always zero?

Think geometrically! The cross product $\hat{a} \times \hat{b}$ is defined to be perpendicular to the plane of $\hat{a}$ and $\hat{b}$. By definition, the dot product of two perpendicular vectors is zero. It is not just a formula; it is the very definition of the cross product's direction.

Q: Can I just find the magnitude of $\vec{c}$ first?

You could, but that would be a trap! Calculating the magnitude of $\vec{c}$ involves squaring the expression, which leads to messy terms like $(\hat{a} \times \hat{b}) \cdot \hat{a}$, which is zero, but also $(\hat{a} \times \hat{b}) \cdot (\hat{a} \times \hat{b})$. It is much faster to distribute the dot product directly into the linear expression.

Select Answer:

Visualized Solution

Visualizing the Vectors $\overset{a}{^}$ and $\hat{b}$

Given unit vectors $\overset{a}{^}$ and $\hat{b}$ , so $∣ \overset{a}{^} ∣ = 1$ and $∣ \hat{b} ∣ = 1$ .
The angle between them is $θ$ , where $sin θ = \frac{65}{9}$ and $0 < θ < \frac{π}{2}$ .

Finding $cos θ$

Using the identity: $cos^{2} θ = 1 - sin^{2} θ$
$cos^{2} θ = 1 - (\frac{65}{9})^{2} = 1 - \frac{65}{81}$
$cos^{2} θ = \frac{81 - 65}{81} = \frac{16}{81}$
Since $0 < θ < \frac{π}{2}$ , $cos θ = \frac{4}{9}$

Defining Vector $c$

Vector $c = 3 \overset{a}{^} + 6 \hat{b} + 9 (\overset{a}{^} \times \hat{b})$
We need to evaluate: $9 (c \cdot \overset{a}{^}) - 3 (c \cdot \hat{b})$

Calculating $c \cdot \overset{a}{^}$ (Expansion)

$c \cdot \overset{a}{^} = (3 \overset{a}{^} + 6 \hat{b} + 9 (\overset{a}{^} \times \hat{b})) \cdot \overset{a}{^}$
Distributing the dot product: $c \cdot \overset{a}{^} = 3 (\overset{a}{^} \cdot \overset{a}{^}) + 6 (\hat{b} \cdot \overset{a}{^}) + 9 ((\overset{a}{^} \times \hat{b}) \cdot \overset{a}{^})$

Simplifying $c \cdot \overset{a}{^}$

Since $\overset{a}{^}$ is a unit vector, $\overset{a}{^} \cdot \overset{a}{^} = 1$ .
The dot product $\hat{b} \cdot \overset{a}{^} = ∣ \hat{b} ∣∣ \overset{a}{^} ∣ cos θ = 1 \cdot 1 \cdot \frac{4}{9} = \frac{4}{9}$ .
The scalar triple product $(\overset{a}{^} \times \hat{b}) \cdot \overset{a}{^} = 0$ because $\overset{a}{^} \times \hat{b}$ is perpendicular to $\overset{a}{^}$ .
$c \cdot \overset{a}{^} = 3 (1) + 6 (\frac{4}{9}) + 0 = 3 + \frac{8}{3} = \frac{17}{3}$

Calculating $c \cdot \hat{b}$ (Expansion)

Now, let's find $c \cdot \hat{b} = (3 \overset{a}{^} + 6 \hat{b} + 9 (\overset{a}{^} \times \hat{b})) \cdot \hat{b}$
Distributing the dot product: $c \cdot \hat{b} = 3 (\overset{a}{^} \cdot \hat{b}) + 6 (\hat{b} \cdot \hat{b}) + 9 ((\overset{a}{^} \times \hat{b}) \cdot \hat{b})$

Simplifying $c \cdot \hat{b}$

We know $\overset{a}{^} \cdot \hat{b} = \frac{4}{9}$ and $\hat{b} \cdot \hat{b} = 1$ .
Again, $(\overset{a}{^} \times \hat{b}) \cdot \hat{b} = 0$ because $\overset{a}{^} \times \hat{b}$ is perpendicular to $\hat{b}$ .
$c \cdot \hat{b} = 3 (\frac{4}{9}) + 6 (1) + 0 = \frac{4}{3} + 6 = \frac{22}{3}$

Final Evaluation

Substitute the calculated values into the target expression: $9 (c \cdot \overset{a}{^}) - 3 (c \cdot \hat{b})$
$9 (\frac{17}{3}) - 3 (\frac{22}{3})$
$= 3 (17) - 22$
$= 51 - 22 = 29$

Conclusion and Key Takeaway

Key Takeaway:
The dot product of a vector with a cross product involving itself is always zero: $(u \times v) \cdot u = 0$ .
This property drastically simplifies complex vector expressions.
Final Answer: $29$

00:00 / 00:00

The Sigma Insight: Scalar (Dot) Product

Solution Diagram

Analyzing the Foundation

We are given two unit vectors,

\overset{a}{^}

and

\hat{b}

, with an angle

θ

between them such that

sin θ = \frac{65}{9}

. Since

\overset{a}{^}

and

\hat{b}

are unit vectors, their magnitudes are

∣ \overset{a}{^} ∣ = 1

and

∣ \hat{b} ∣ = 1

To find the dot product

\overset{a}{^} \cdot \hat{b} = cos θ

, we use the trigonometric identity

cos^{2} θ = 1 - sin^{2} θ

. Substituting the given value:

cos^{2} θ = 1 - \frac{65}{81} = \frac{16}{81}

Assuming

θ

is acute, we take the positive root to obtain:

cos θ = \frac{4}{9}

The Deconstruction of $c$

The vector

c

is defined as:

c = 3 \overset{a}{^} + 6 \hat{b} + 9 (\overset{a}{^} \times \hat{b})

When calculating the dot product

c \cdot \overset{a}{^}

, we distribute the operation across the sum:

c \cdot \overset{a}{^} = 3 (\overset{a}{^} \cdot \overset{a}{^}) + 6 (\hat{b} \cdot \overset{a}{^}) + 9 ((\overset{a}{^} \times \hat{b}) \cdot \overset{a}{^})

Because

\overset{a}{^} \cdot \overset{a}{^} = 1

and the scalar triple product

(\overset{a}{^} \times \hat{b}) \cdot \overset{a}{^} = 0

(due to the orthogonality of the cross product), the expression simplifies to:

c \cdot \overset{a}{^} = 3 (1) + 6 (\frac{4}{9}) = 3 + \frac{8}{3} = \frac{17}{3}

Calculating the Second Component

Next, we evaluate

c \cdot \hat{b}

using the same distributive property:

c \cdot \hat{b} = 3 (\overset{a}{^} \cdot \hat{b}) + 6 (\hat{b} \cdot \hat{b}) + 9 ((\overset{a}{^} \times \hat{b}) \cdot \hat{b})

Again, the term

(\overset{a}{^} \times \hat{b}) \cdot \hat{b}

vanishes because the cross product is perpendicular to

\hat{b}

. We calculate the remaining terms:

c \cdot \hat{b} = 3 (\frac{4}{9}) + 6 (1) = \frac{4}{3} + 6 = \frac{22}{3}

Final Calculation

We now substitute these results into the target expression

9 (c \cdot \overset{a}{^}) - 3 (c \cdot \hat{b})

9 (\frac{17}{3}) - 3 (\frac{22}{3})

Simplifying the arithmetic:

3 (17) - 22 = 51 - 22 = 29

The final value of the expression is 29.

Common Traps & FAQs

Wait, why do we reject the negative root for

cos θ

Great catch! The problem explicitly states

0 < θ < \frac{π}{2}

. In the first quadrant, the cosine function is strictly positive. If we had chosen the negative root, we would be describing an obtuse angle, which contradicts the given constraints.

Why is the dot product of the cross product with its own components always zero?

Think geometrically! The cross product

\overset{a}{^} \times \hat{b}

is defined to be perpendicular to the plane of

\overset{a}{^}

and

\hat{b}

. By definition, the dot product of two perpendicular vectors is zero. It is not just a formula; it is the very definition of the cross product's direction.

Can I just find the magnitude of

c

first?

You could, but that would be a trap! Calculating the magnitude of

c

involves squaring the expression, which leads to messy terms like

(\overset{a}{^} \times \hat{b}) \cdot \overset{a}{^}

, which is zero, but also

(\overset{a}{^} \times \hat{b}) \cdot (\overset{a}{^} \times \hat{b})

. It is much faster to distribute the dot product directly into the linear expression.

Select Answer:

Visualized Solution

Visualizing the Vectors $\overset{a}{^}$ and $\hat{b}$

Finding $cos θ$

Defining Vector $c$

Calculating $c \cdot \overset{a}{^}$ (Expansion)

Simplifying $c \cdot \overset{a}{^}$

Calculating $c \cdot \hat{b}$ (Expansion)

Simplifying $c \cdot \hat{b}$

Final Evaluation

Conclusion and Key Takeaway

The Sigma Insight: Scalar (Dot) Product

Analyzing the Foundation

The Deconstruction of $c$

Calculating the Second Component

Final Calculation

Similar Questions

Let a unit vector $C$ which makes an angle of $6 0^{\circ}$ with $2 \hat{i} + 2 \hat{j} - \hat{k}$ and angle $4 5^{\circ}$ with $\hat{i} - \hat{k}$ be $C$ . Then $C + (- \frac{1}{2} \hat{i} + \frac{1}{3 2} \hat{j} - \frac{2}{3} \hat{k})$ is :

Let $a$ and $b$ be two vectors such that $∣2 a + 3 b ∣ = ∣3 a + b ∣$ and the angle between $a$ and $b$ is $6 0^{\circ}$ . If $\frac{1}{8} a$ is a unit vector, then $∣ b ∣$ is equal to :

If $a, b, c$ are vectors show that $a + b + c = 0$ and $∣ a ∣ = 7, ∣ b ∣ = 5, ∣ c ∣ = 3$ then angle between vector $b$ and $c$ is

Let $a, b, c \in R$ be such that $a^{2} + b^{2} + c^{2} = 1$ . If $a cos θ = b cos (θ + \frac{2 π}{3}) = c cos (θ + \frac{4 π}{3})$ , where $θ = \frac{π}{9}$ , then the angle between the vectors $a \hat{i} + b \hat{j} + c \hat{k}$ and $b \hat{i} + c \hat{j} + a \hat{k}$ is :

Let $a$ and $b$ be two unit vectors. If the vectors $c = a + 2 b$ and $d = 5 a - 4 b$ are perpendicular to each other, then the angle between $a$ and $b$ is:

If $a$ and $b$ are two unit vectors such that $a + 2 b$ and $5 a - 4 b$ are perpendicular to each other then the angle between $a$ and $b$ is

Let $a = \hat{i} + 2 \hat{j} + 3 \hat{k}, b = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ and $c = 3 \hat{i} - \hat{j} + λ \hat{k}$ be three vectors. Let $r$ be a unit vector along $b + c$ . If $r \cdot a = 3$ , then $3 λ$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a^ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b^

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosθ

Defining Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c⋅a^ (Expansion)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c⋅a^

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c⋅b^ (Expansion)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c⋅b^

Final Evaluation

Conclusion and Key Takeaway

The Sigma Insight: Scalar (Dot) Product

Analyzing the Foundation

The Deconstruction of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Calculating the Second Component

Final Calculation

Similar Questions

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Visualizing the Vectors $\overset{a}{^}$ and $\hat{b}$

Finding $cos θ$

Defining Vector $c$

Calculating $c \cdot \overset{a}{^}$ (Expansion)

Simplifying $c \cdot \overset{a}{^}$

Calculating $c \cdot \hat{b}$ (Expansion)

Simplifying $c \cdot \hat{b}$

The Deconstruction of $c$

Let a unit vector $C$ which makes an angle of $6 0^{\circ}$ with $2 \hat{i} + 2 \hat{j} - \hat{k}$ and angle $4 5^{\circ}$ with $\hat{i} - \hat{k}$ be $C$ . Then $C + (- \frac{1}{2} \hat{i} + \frac{1}{3 2} \hat{j} - \frac{2}{3} \hat{k})$ is :

Let $a$ and $b$ be two vectors such that $∣2 a + 3 b ∣ = ∣3 a + b ∣$ and the angle between $a$ and $b$ is $6 0^{\circ}$ . If $\frac{1}{8} a$ is a unit vector, then $∣ b ∣$ is equal to :

If $a, b, c$ are vectors show that $a + b + c = 0$ and $∣ a ∣ = 7, ∣ b ∣ = 5, ∣ c ∣ = 3$ then angle between vector $b$ and $c$ is

Let $a, b, c \in R$ be such that $a^{2} + b^{2} + c^{2} = 1$ . If $a cos θ = b cos (θ + \frac{2 π}{3}) = c cos (θ + \frac{4 π}{3})$ , where $θ = \frac{π}{9}$ , then the angle between the vectors $a \hat{i} + b \hat{j} + c \hat{k}$ and $b \hat{i} + c \hat{j} + a \hat{k}$ is :

Let $a$ and $b$ be two unit vectors. If the vectors $c = a + 2 b$ and $d = 5 a - 4 b$ are perpendicular to each other, then the angle between $a$ and $b$ is:

If $a$ and $b$ are two unit vectors such that $a + 2 b$ and $5 a - 4 b$ are perpendicular to each other then the angle between $a$ and $b$ is

Let $a = \hat{i} + 2 \hat{j} + 3 \hat{k}, b = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ and $c = 3 \hat{i} - \hat{j} + λ \hat{k}$ be three vectors. Let $r$ be a unit vector along $b + c$ . If $r \cdot a = 3$ , then $3 λ$ is equal to: