Let $\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$ and $\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\overrightarrow a$ and $\overrightarrow b$ is $8\sqrt 3$ square units, then $\overrightarrow a$ . $\overrightarrow b$ is equal to __________.
Answer (integer)
2
Solution
$\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$<br><br>$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$<br><br>Area of parallelogram = $\left| {\overrightarrow a \times \overrightarrow b } \right|$<br><br>$$ = \left| {(\widehat i + \alpha \widehat j + 3\widehat k) \times (3\widehat i - \alpha \widehat j + \widehat k)} \right|$$<br><br>$$8\sqrt 3 = \left| {(4\alpha )\widehat i + 8\widehat j - (4\alpha )\widehat k} \right|$$<br><br>$(64)(3) = 16{\alpha ^2} + 64 + 16{\alpha ^2}$<br><br>$(64)(3) = 32{\alpha ^2} + 64$<br><br>$6 = {\alpha ^2} + 2$<br><br>${\alpha ^2} = 4$<br><br>$\therefore$ $\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$<br><br>$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$<br><br>$\overrightarrow a \,.\,\overrightarrow b = 3 - {\alpha ^2} + 3$<br><br>$= 6 - {\alpha ^2}$<br><br>$= 6 - 4$<br><br>$= 2$
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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