Key Concepts and Formulas

• Direction Cosines: For a vector $\overrightarrow v = x\widehat i + y\widehat j + z\widehat k$, the direction cosines are $\cos \alpha = \frac{x}{|\overrightarrow v|}$, $\cos \beta = \frac{y}{|\overrightarrow v|}$, and $\cos \gamma = \frac{z}{|\overrightarrow v|}$. The property $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$ holds.
• Vector Projection: The scalar projection of vector $\overrightarrow A$ onto vector $\overrightarrow B$ is $\frac{\overrightarrow A \cdot \overrightarrow B}{|\overrightarrow B|}$.
• Dot Product: $\overrightarrow A \cdot \overrightarrow B = |\overrightarrow A| |\overrightarrow B| \cos \theta$. If $\overrightarrow A \perp \overrightarrow B$, then $\overrightarrow A \cdot \overrightarrow B = 0$.
• Coplanarity: Three vectors $\overrightarrow A, \overrightarrow B, \overrightarrow C$ are coplanar if their scalar triple product $[\overrightarrow A \overrightarrow B \overrightarrow C] = 0$.

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Step-by-Step Solution

Step 1: Determine the vector $\overrightarrow a$.

We are given that $\overrightarrow a = a_1\widehat i + a_2\widehat j + a_3\widehat k$ with $a_i > 0$. The vector $\overrightarrow a$ makes equal angles with the coordinate axes. Let these angles be $\alpha, \beta, \gamma$. So, $\alpha = \beta = \gamma$. This implies their cosines are equal: $\cos \alpha = \cos \beta = \cos \gamma$. Let this common value be $c$. Using the property of direction cosines, $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Substituting $c$ for each cosine: $c^2 + c^2 + c^2 = 1 \implies 3c^2 = 1 \implies c^2 = \frac{1}{3}$. Thus, $c = \pm \frac{1}{\sqrt{3}}$. Since $a_i > 0$, the vector lies in the first octant, meaning the angles with the positive axes are acute. Therefore, $\cos \alpha > 0$, so $c = \frac{1}{\sqrt{3}}$. The direction cosines are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$. Let $|\overrightarrow a| = \lambda$. Then, $\overrightarrow a = \lambda \left( \frac{1}{\sqrt{3}}\widehat i + \frac{1}{\sqrt{3}}\widehat j + \frac{1}{\sqrt{3}}\widehat k \right) = \frac{\lambda}{\sqrt{3}}(\widehat i + \widehat j + \widehat k)$ We are given that the projection of $\overrightarrow a$ on $\overrightarrow v = 3\widehat i + 4\widehat j$ is 7. The magnitude of $\overrightarrow v$ is $|\overrightarrow v| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. The dot product $\overrightarrow a \cdot \overrightarrow v$ is: $\overrightarrow a \cdot \overrightarrow v = \left( \frac{\lambda}{\sqrt{3}}(\widehat i + \widehat j + \widehat k) \right) \cdot (3\widehat i + 4\widehat j) = \frac{\lambda}{\sqrt{3}}(1 \cdot 3 + 1 \cdot 4 + 1 \cdot 0) = \frac{7\lambda}{\sqrt{3}}$ The projection is $\frac{\overrightarrow a \cdot \overrightarrow v}{|\overrightarrow v|} = \frac{7\lambda/\sqrt{3}}{5} = \frac{7\lambda}{5\sqrt{3}}$. We are given this is equal to 7: $\frac{7\lambda}{5\sqrt{3}} = 7 \implies \lambda = 5\sqrt{3}$ Substituting $\lambda$ back into $\overrightarrow a$: $\overrightarrow a = \frac{5\sqrt{3}}{\sqrt{3}}(\widehat i + \widehat j + \widehat k) = 5(\widehat i + \widehat j + \widehat k)$

Step 2: Determine the properties of vector $\overrightarrow b$.

Vector $\overrightarrow b$ is obtained by rotating $\overrightarrow a$ by $90^\circ$. This implies two conditions:

1. $|\overrightarrow b| = |\overrightarrow a| = 5\sqrt{3}$.
2. $\overrightarrow a \cdot \overrightarrow b = 0$ (since they are perpendicular after a $90^\circ$ rotation).

Let $\overrightarrow b = p\widehat i + q\widehat j + r\widehat k$. From $\overrightarrow a \cdot \overrightarrow b = 0$: $5(p) + 5(q) + 5(r) = 0 \implies p + q + r = 0 \quad \text{... (1)}$ We are also given that $\overrightarrow a$, $\overrightarrow b$, and the x-axis (represented by $\widehat i$) are coplanar. The scalar triple product $[\overrightarrow a \overrightarrow b \widehat i]$ must be zero: $\left| \begin{matrix} 5 & 5 & 5 \\ p & q & r \\ 1 & 0 & 0 \end{matrix} \right| = 0$ Expanding the determinant along the third row: $1 \cdot (5r - 5q) = 0 \implies 5r - 5q = 0 \implies r = q \quad \text{... (2)}$

Step 3: Find the components of $\overrightarrow b$.

Substitute $r=q$ from (2) into (1): $p + q + q = 0 \implies p + 2q = 0 \implies p = -2q$ So, $\overrightarrow b$ can be written as: $\overrightarrow b = (-2q)\widehat i + q\widehat j + q\widehat k = q(-2\widehat i + \widehat j + \widehat k)$ Now, use the magnitude condition $|\overrightarrow b| = 5\sqrt{3}$: $|q(-2\widehat i + \widehat j + \widehat k)| = 5\sqrt{3}$ $|q| \sqrt{(-2)^2 + 1^2 + 1^2} = 5\sqrt{3}$ $|q| \sqrt{4 + 1 + 1} = 5\sqrt{3}$ $|q| \sqrt{6} = 5\sqrt{3}$ $|q| = \frac{5\sqrt{3}}{\sqrt{6}} = \frac{5}{\sqrt{2}}$ Let's choose $q = \frac{5}{\sqrt{2}}$ (the sign of $q$ will only affect the sign of the projection, not its magnitude). Then, $\overrightarrow b = \frac{5}{\sqrt{2}}(-2\widehat i + \widehat j + \widehat k)$.

Step 4: Calculate the projection of $\overrightarrow b$ on $3\widehat i + 4\widehat j$.

Let $\overrightarrow u = 3\widehat i + 4\widehat j$. We need to find the scalar projection of $\overrightarrow b$ on $\overrightarrow u$, which is $\frac{\overrightarrow b \cdot \overrightarrow u}{|\overrightarrow u|}$. We know $|\overrightarrow u| = 5$. Now, compute the dot product $\overrightarrow b \cdot \overrightarrow u$: $\overrightarrow b \cdot \overrightarrow u = \left( \frac{5}{\sqrt{2}}(-2\widehat i + \widehat j + \widehat k) \right) \cdot (3\widehat i + 4\widehat j)$ $\overrightarrow b \cdot \overrightarrow u = \frac{5}{\sqrt{2}}((-2)(3) + (1)(4) + (1)(0))$ $\overrightarrow b \cdot \overrightarrow u = \frac{5}{\sqrt{2}}(-6 + 4) = \frac{5}{\sqrt{2}}(-2) = -\frac{10}{\sqrt{2}} = -5\sqrt{2}$ The scalar projection is $\frac{-5\sqrt{2}}{5} = -\sqrt{2}$. The question asks for "the projection", and given the options are positive, it implies the length of the projection. Length of projection $= |-\sqrt{2}| = \sqrt{2}$.

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Common Mistakes & Tips

• Direction Cosines: Remember that $a_i > 0$ implies the vector is in the first octant, so its direction cosines must be positive.
• Projection: Be mindful whether the question asks for the scalar projection (which can be negative) or the length of the projection (which is always non-negative). The options provided suggest the length is required.
• Coplanarity: The determinant calculation for coplanarity is a key step; ensure it's done correctly, especially when expanding along rows/columns with zeros.

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Summary

We began by determining the vector $\overrightarrow a$ using the equal angle condition and the given projection value. Subsequently, we leveraged the properties of vector rotation (equal magnitude and perpendicularity) and the coplanarity condition with the x-axis to establish relationships between the components of $\overrightarrow b$. Solving these equations allowed us to find the explicit form of $\overrightarrow b$. Finally, we calculated the projection of $\overrightarrow b$ onto the vector $3\widehat i + 4\widehat j$, yielding a length of $\sqrt{2}$.

The final answer is $\boxed{\sqrt 2}$.