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Step-by-Step Solution
Step 1: Identify the given vectors
Let
\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k} \,.
We are given two vectors:
\overrightarrow{u} = 2\hat{i} + 4\hat{j} - 5\hat{k} \quad \text{and} \quad \overrightarrow{v} = -\lambda \hat{i} + 2\hat{j} + 3\hat{k} \,.
The sum of these two vectors is
\overrightarrow{b} = \overrightarrow{u} + \overrightarrow{v} = (2 - \lambda)\hat{i} + (4 + 2)\hat{j} + (-5 + 3)\hat{k} = (2 - \lambda)\hat{i} + 6\hat{j} - 2\hat{k} \,.
Step 2: Write down the formula for the projection of \overrightarrow{a} on \overrightarrow{b}
The projection of \overrightarrow{a} on \overrightarrow{b} (in terms of magnitude) is given by
\displaystyle \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{\lvert \overrightarrow{b} \rvert} \,.
We are told this projection equals 1:
\displaystyle \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{\lvert \overrightarrow{b} \rvert} = 1 \,.
Step 3: Compute \overrightarrow{a} \cdot \overrightarrow{b}
Using the dot product definition:
\overrightarrow{a} \cdot \overrightarrow{b} = (\hat{i} + 2\hat{j} + \hat{k}) \cdot \bigl((2-\lambda)\hat{i} + 6\hat{j} - 2\hat{k}\bigr).
This expands to
\bigl(1 \times (2 - \lambda)\bigr) + \bigl(2 \times 6\bigr) + \bigl(1 \times (-2)\bigr) \\
= (2 - \lambda) + 12 - 2 \\
= (2 - \lambda) + 10 \\
= 12 - \lambda \,.
Step 4: Compute \lvert \overrightarrow{b} \rvert
First, find \lvert \overrightarrow{b} \rvert^2 :
\lvert \overrightarrow{b} \rvert^2
= (2-\lambda)^2 + 6^2 + (-2)^2
= (2-\lambda)^2 + 36 + 4
= (2-\lambda)^2 + 40 \,.
Thus,
\lvert \overrightarrow{b} \rvert = \sqrt{(2-\lambda)^2 + 40} \,.
Step 5: Apply the projection condition
We have
\displaystyle \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{\lvert \overrightarrow{b} \rvert} = 1 \,.
Substitute the expressions obtained:
\displaystyle \frac{12 - \lambda}{\sqrt{(2-\lambda)^2 + 40}} = 1 \,.
This implies
12 - \lambda = \sqrt{(2-\lambda)^2 + 40} \,.
Step 6: Solve for \lambda
Square both sides:
(12 - \lambda)^2 = (2-\lambda)^2 + 40 \,.
Expand:
144 - 24\lambda + \lambda^2 = (2 - \lambda)^2 + 40 \\
144 - 24\lambda + \lambda^2 = (4 - 4\lambda + \lambda^2) + 40 \\
144 - 24\lambda + \lambda^2 = \lambda^2 - 4\lambda + 44 \,.
Cancel \lambda^2 on both sides:
144 - 24\lambda = -4\lambda + 44 \,.
Bring like terms together:
144 - 44 = 24\lambda - 4\lambda \\
100 = 20\lambda \\
\lambda = 5 \,.
Step 7: Conclude the value of \lambda
Hence, the required value of \lambda is
\boxed{5} \,.
