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Step-by-Step Solution
Step 1: Determine (p + q) and (p − q)
Given:
\overrightarrow{p} = 2\widehat{i} + 3\widehat{j} + \widehat{k},
\quad
\overrightarrow{q} = \widehat{i} + 2\widehat{j} + \widehat{k}.
First, find their sums and differences:
\overrightarrow{p} + \overrightarrow{q} = (2+1)\widehat{i} + (3+2)\widehat{j} + (1+1)\widehat{k}
= 3\widehat{i} + 5\widehat{j} + 2\widehat{k},
\overrightarrow{p} - \overrightarrow{q} = (2-1)\widehat{i} + (3-2)\widehat{j} + (1-1)\widehat{k}
= 1\widehat{i} + 1\widehat{j} + 0\widehat{k}.
Step 2: Express r as a vector perpendicular to both (p + q) and (p − q)
A vector perpendicular to two given vectors is parallel to their cross product. Thus, if
\overrightarrow{r}
= \alpha \widehat{i} + \beta \widehat{j} + \gamma \widehat{k}
is perpendicular to both
(\overrightarrow{p} + \overrightarrow{q}) and
(\overrightarrow{p} - \overrightarrow{q}) ,
then
\overrightarrow{r} \propto (\overrightarrow{p} + \overrightarrow{q})
\times (\overrightarrow{p} - \overrightarrow{q}).
Step 3: Compute the cross product (p + q) × (p − q)
(\overrightarrow{p} + \overrightarrow{q})
\times (\overrightarrow{p} - \overrightarrow{q})
= \begin{vmatrix}
\widehat{i} & \widehat{j} & \widehat{k} \\
3 & 5 & 2 \\
1 & 1 & 0
\end{vmatrix}.
To find this determinant:
= \widehat{i}(5\cdot0 - 2\cdot1)
- \widehat{j}(3\cdot0 - 2\cdot1)
+ \widehat{k}(3\cdot1 - 5\cdot1)
= \widehat{i}(0 - 2)
- \widehat{j}(0 - 2)
+ \widehat{k}(3 - 5)
= -2\widehat{i} + 2\widehat{j} - 2\widehat{k}.
(Note: Some sign manipulations might lead to equivalent forms; the important point is that
\overrightarrow{r} is parallel to this vector.)
Step 4: Find the exact form of r using the magnitude condition
Since
\overrightarrow{r}
must be parallel to
-2\widehat{i} + 2\widehat{j} - 2\widehat{k} ,
we can write
\overrightarrow{r} = \lambda(-2\widehat{i} + 2\widehat{j} - 2\widehat{k}),
for some scalar \lambda . The question also states that
|\overrightarrow{r}| = \sqrt{3}.
Step 5: Determine λ using the given magnitude
First, compute the magnitude of the vector
-2\widehat{i} + 2\widehat{j} - 2\widehat{k} :
\sqrt{(-2)^2 + (2)^2 + (-2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}.
Thus,
|\overrightarrow{r}|
= |\lambda|\cdot 2\sqrt{3}
= \sqrt{3}.
This implies
|\lambda| \cdot 2\sqrt{3} = \sqrt{3}
\quad\Rightarrow\quad
|\lambda| = \frac{1}{2}.
Hence,
\lambda = \pm \frac{1}{2}.
Step 6: Write down the components of r
Substituting \lambda = \pm \frac{1}{2} ,
\overrightarrow{r}
= \pm \frac{1}{2}\left(-2\widehat{i} + 2\widehat{j} - 2\widehat{k}\right)
= \pm\left(-\widehat{i} + \widehat{j} - \widehat{k}\right).
Therefore,
\alpha = \pm(-1),
\ \beta = \pm(1),
\ \gamma = \pm(-1).
In absolute values,
|\alpha| = 1,\ |\beta| = 1,\ |\gamma| = 1.
Step 7: Calculate |α| + |β| + |γ|
Since each of |\alpha| , |\beta| , and |\gamma| equals 1,
|\alpha| + |\beta| + |\gamma| = 1 + 1 + 1 = 3.
Final Answer
The value of |\alpha| + |\beta| + |\gamma| is 3.
