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Step-by-Step Solution
Step 1: Identify the given vectors and known conditions
• We have three vectors:
\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k} ,
\overrightarrow{b} (unknown),
\overrightarrow{c} = \hat{j} - \hat{k} .
• It is given that:
1. \overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c} .
2. \overrightarrow{a} \cdot \overrightarrow{b} = 1 .
• We want to find 3 \, l^2 , where l is the length of the projection of \overrightarrow{b} on \overrightarrow{a} \times \overrightarrow{c} .
Step 2: Compute the magnitude of vector \overrightarrow{c}
Since \overrightarrow{c} = \hat{j} - \hat{k} ,
its magnitude squared is:
|\overrightarrow{c}|^2 = (0)^2 + (1)^2 + (-1)^2 = 1 + 1 = 2 .
Thus, |\overrightarrow{c}| = \sqrt{2} .
Step 3: Use the condition \overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c}
Taking the dot product of both sides with \overrightarrow{c} , we get:
( \overrightarrow{a} \times \overrightarrow{b} ) \cdot \overrightarrow{c} = \overrightarrow{c} \cdot \overrightarrow{c} = |\overrightarrow{c}|^2 = 2.
Step 4: Relate \overrightarrow{b} and its magnitude using known vector identities
From the formula for the magnitude of a cross product, we have:
|\overrightarrow{a} \times \overrightarrow{b}|^2 = |\overrightarrow{a}|^2 \,|\overrightarrow{b}|^2 - (\overrightarrow{a} \cdot \overrightarrow{b})^2.
Here, |\overrightarrow{a} \times \overrightarrow{b}|^2 = |\overrightarrow{c}|^2 = 2 . Also, |\overrightarrow{a}|^2 = 1^2 + 1^2 + 1^2 = 3 , and \overrightarrow{a} \cdot \overrightarrow{b} = 1 . Substituting these:
2 = 3 \, |\overrightarrow{b}|^2 - (1)^2,
which gives 2 = 3 \, |\overrightarrow{b}|^2 - 1
\implies 3 \, |\overrightarrow{b}|^2 = 3 \implies |\overrightarrow{b}|^2 = 1 \implies |\overrightarrow{b}| = 1.
Step 5: Compute \overrightarrow{a} \times \overrightarrow{c}
We can use the determinant form to find \overrightarrow{a} \times \overrightarrow{c} :
\overrightarrow{a} \times \overrightarrow{c}
=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
0 & 1 & -1
\end{vmatrix}
= \hat{i} (1 \cdot (-1) - 1 \cdot 1)
- \hat{j} (1 \cdot (-1) - 1 \cdot 0)
+ \hat{k} (1 \cdot 1 - 1 \cdot 0).
Simplifying this:
= \hat{i} (-1 - 1)
- \hat{j} (-1 - 0)
+ \hat{k} (1 - 0)
= -2 \hat{i} + \hat{j} + \hat{k}.
Hence,
\overrightarrow{a} \times \overrightarrow{c} = -2 \hat{i} + \hat{j} + \hat{k},
and its magnitude is
| \overrightarrow{a} \times \overrightarrow{c} | = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}.
Step 6: Find the scalar \overrightarrow{b} \cdot ( \overrightarrow{a} \times \overrightarrow{c} )
Using the vector identity
\overrightarrow{a} \times ( \overrightarrow{a} \times \overrightarrow{b} ) = \overrightarrow{a} \, (\overrightarrow{a} \cdot \overrightarrow{b}) - \overrightarrow{b} \, |\overrightarrow{a}|^2,
we substitute \overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c} to get:
\overrightarrow{a} \times \overrightarrow{c}
= \overrightarrow{a} \times ( \overrightarrow{a} \times \overrightarrow{b} )
= \overrightarrow{a} ( \overrightarrow{a} \cdot \overrightarrow{b} )
- \overrightarrow{b} ( |\overrightarrow{a}|^2 ).
Because \overrightarrow{a} \cdot \overrightarrow{b} = 1 and |\overrightarrow{a}|^2 = 3 , we have:
\overrightarrow{a} \times \overrightarrow{c}
= \overrightarrow{a} (1) - \overrightarrow{b} (3)
= \overrightarrow{a} - 3 \, \overrightarrow{b}.
Taking the dot product with \overrightarrow{b} on both sides:
\overrightarrow{b} \cdot ( \overrightarrow{a} \times \overrightarrow{c} )
= \overrightarrow{b} \cdot (\overrightarrow{a} - 3 \, \overrightarrow{b} )
= \underbrace{\overrightarrow{b} \cdot \overrightarrow{a}}_{=1}
- 3 \underbrace{\overrightarrow{b} \cdot \overrightarrow{b}}_{|\overrightarrow{b}|^2=1}
= 1 - 3(1) = -2.
Thus, \overrightarrow{b} \cdot ( \overrightarrow{a} \times \overrightarrow{c} ) = -2. The absolute value will be 2.
Step 7: Determine the projection of \overrightarrow{b} on \overrightarrow{a} \times \overrightarrow{c}
The length of the projection of \overrightarrow{b} onto \overrightarrow{a} \times \overrightarrow{c} is given by:
l
= \frac{ \big| \overrightarrow{b} \cdot (\overrightarrow{a} \times \overrightarrow{c}) \big| }
{ \big| \overrightarrow{a} \times \overrightarrow{c} \big| }
= \frac{ 2 }{ \sqrt{6} }.
Hence,
l^2 = \left( \frac{2}{\sqrt{6}} \right)^2 = \frac{4}{6} = \frac{2}{3}.
Step 8: Compute 3 \, l^2
We have:
3 \, l^2 = 3 \times \frac{2}{3} = 2.
Therefore, the value of 3 \, l^2 is \boxed{2}.
