If the vectors, $$\overrightarrow p = \left( {a + 1} \right)\widehat i + a\widehat j + a\widehat k$$,
$$\overrightarrow q = a\widehat i + \left( {a + 1} \right)\widehat j + a\widehat k$$ and
$$\overrightarrow r = a\widehat i + a\widehat j + \left( {a + 1} \right)\widehat k\left( {a \in R} \right)$$
are coplanar
and $$3{\left( {\overrightarrow p .\overrightarrow q } \right)^2} - \lambda \left| {\overrightarrow r \times \overrightarrow q } \right|^2 = 0$$, then the value of $\lambda$ is ______.
Answer (integer)
1
Solution
$\because$ $\overrightarrow p$, $\overrightarrow q$, $\overrightarrow r$ are coplanar
<br><br>$\therefore$ $$\left[ {\matrix{
{\overrightarrow p } & {\overrightarrow q } & {\overrightarrow r } \cr
} } \right]$$ = 0
<br><br>$\Rightarrow$ $$\left| {\matrix{
{a + 1} & a & a \cr
a & {a + 1} & a \cr
a & a & {a + 1} \cr
} } \right|$$ = 0
<br><br>$\Rightarrow$ (a + 1) + a + a = 0
<br><br>$\Rightarrow$ a = $- {1 \over 3}$
<br><br>$\overrightarrow p .\overrightarrow q$ = ${1 \over 9}\left( { - 2 - 2 + 1} \right)$ = $- {1 \over 3}$
<br><br>$\overrightarrow r \times \overrightarrow q$ = $${1 \over 9}\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
{ - 1} & 2 & { - 1} \cr
{ - 1} & { - 1} & 2 \cr
} } \right|$$
<br><br>= ${1 \over 9}\left( {3\widehat i + 3\widehat j + 3\widehat k} \right)$
<br><br>= ${{\widehat i + \widehat j + \widehat k} \over 3}$
<br><br>$\Rightarrow$ ${\left| {\overrightarrow r \times \overrightarrow q } \right|^2} = {1 \over 3}$
<br><br>Also $$3{\left( {\overrightarrow p .\overrightarrow q } \right)^2} - \lambda \left| {\overrightarrow r \times \overrightarrow q } \right|^2 = 0$$
<br><br>$\Rightarrow$ $3\left( {{1 \over 9}} \right) - \lambda \left( {{1 \over 3}} \right)$ = 0
<br><br> $\Rightarrow$ $\lambda$ = 1
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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