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Step-by-Step Solution
Step 1: Identify the given vectors
We have the following vectors:
• \vec{a} = 2\hat{i} \;-\;\hat{j} \;+\;2\hat{k}
• \vec{b} = \hat{i} \;+\;2\hat{j} \;-\;\hat{k}
• \vec{c} = 3\hat{i} \;+\;2\hat{j} \;-\;\hat{k}
Step 2: Express \vec{v} in the plane of \vec{a} and \vec{b}
Since \vec{v} lies in the plane spanned by \vec{a} and \vec{b} , we can write
\vec{v} = x\,\vec{a} \;+\; y\,\vec{b}
for some scalars x and y .
Step 3: Use the condition that \vec{v} is perpendicular to \vec{c}
We are told that \vec{v} \perp \vec{c} . Hence their dot product is zero:
\vec{v} \cdot \vec{c} \;=\; 0.
A useful identity to handle this condition is:
\vec{v} \;=\; \lambda \bigl[\vec{c} \times (\vec{a} \times \vec{b})\bigr],
for some scalar \lambda . This ensures automatically that \vec{v} \cdot \vec{c} = 0 .
Step 4: Rewrite \vec{c} \times (\vec{a} \times \vec{b}) using vector identities
Using the vector triple product formula
\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\,\vec{B} - (\vec{A} \cdot \vec{B})\,\vec{C},
we get
\vec{c} \times (\vec{a} \times \vec{b})
=
\bigl[(\vec{c} \cdot \vec{b})\,\vec{a} - (\vec{c} \cdot \vec{a})\,\vec{b}\bigr].
Let us compute these dot products first:
1. \vec{c} \cdot \vec{b} \;=\; (3)(1) + (2)(2) + (-1)(-1) \;=\; 3 \;+\; 4 \;+\; 1 \;=\; 8.
2. \vec{c} \cdot \vec{a} \;=\; (3)(2) + (2)(-1) + (-1)(2) \;=\; 6 \;-\; 2 \;-\; 2 \;=\; 2.
Hence,
\vec{c} \times (\vec{a} \times \vec{b})
= 8\vec{a} \;-\; 2\vec{b}.
Therefore,
\vec{v}
= \lambda \Bigl[\,8\,\vec{a} \;-\; 2\,\vec{b}\Bigr].
Step 5: Apply the projection condition on \vec{a}
We are told the projection of \vec{v} on \vec{a} is 19 units. The projection of a vector \vec{v} on \vec{a} is given by
\text{proj}_{\vec{a}}(\vec{v})
= \dfrac{\vec{v} \cdot \vec{a}}{\lvert \vec{a} \rvert}.
Thus, the condition is
\dfrac{\vec{v} \cdot \vec{a}}{\lvert \vec{a} \rvert} = 19.
Let us compute \vec{v} \cdot \vec{a} using
\vec{v} = \lambda\,(8\,\vec{a} - 2\,\vec{b}).
\vec{v} \cdot \vec{a}
= \lambda\bigl(8\,\vec{a} \cdot \vec{a} - 2\,\vec{b} \cdot \vec{a}\bigr).
Next, compute the needed dot products:
• \vec{a} \cdot \vec{a} = (2)^2 + (-1)^2 + (2)^2 = 4 + 1 + 4 = 9.
• \vec{b} \cdot \vec{a}
= (1)(2) + (2)(-1) + (-1)(2)
= 2 - 2 - 2 = -2.
Hence,
\vec{v} \cdot \vec{a}
= \lambda \bigl(8 \times 9 - 2 \times (-2)\bigr)
= \lambda \bigl(72 + 4\bigr)
= 76\,\lambda.
Also, \lvert \vec{a} \rvert = \sqrt{9} = 3. Thus the projection condition
\frac{\vec{v} \cdot \vec{a}}{\lvert \vec{a} \rvert} = 19
becomes
\frac{76\,\lambda}{3} = 19.
Solving for \lambda :
76\,\lambda = 19 \times 3 \quad\Longrightarrow\quad 76\,\lambda = 57 \quad\Longrightarrow\quad \lambda = \frac{57}{76} = \frac{3}{4}.
Step 6: Determine \vec{v}
Substituting \lambda = \tfrac{3}{4} back, we get
\vec{v}
= \frac{3}{4}\,\Bigl[\,8\,\vec{a} \;-\; 2\,\vec{b}\Bigr]
= 6\,\vec{a} \;-\; \frac{3}{2}\,\vec{b}.
Step 7: Compute |2\vec{v}|^2
First, note that
2\vec{v} = 2\Bigl(\frac{3}{4}[\,8\,\vec{a} - 2\,\vec{b}]\Bigr)
= \frac{3}{2}[\,8\,\vec{a} - 2\,\vec{b}]
= 12\,\vec{a} - 3\,\vec{b}.
We want |2\vec{v}|^2 = (2\vec{v}) \cdot (2\vec{v}).
Substitute the expression in:
2\vec{v} = 12\,\vec{a} - 3\,\vec{b}.
So,
|2\vec{v}|^2
= (12\,\vec{a} - 3\,\vec{b}) \cdot (12\,\vec{a} - 3\,\vec{b}).
Expand the dot product:
|2\vec{v}|^2
= 144\,(\vec{a} \cdot \vec{a})
\;-\; 72\,(\vec{a} \cdot \vec{b})
\;+\; 9\,(\vec{b} \cdot \vec{b}).
We have already computed or can compute:
1. \vec{a} \cdot \vec{a} = 9.
2. \vec{a} \cdot \vec{b} = -2.
3. \vec{b} \cdot \vec{b} = (1)^2 + (2)^2 + (-1)^2 = 1 + 4 + 1 = 6.
Thus,
|2\vec{v}|^2
= 144 \times 9 \;-\; 72 \times (-2) \;+\; 9 \times 6
= 1296 + 144 + 54
= 1494.
Final Answer
|2\vec{v}|^2 = 1494.
