First slide

Algebra of vectors

Question

Suppose that $\vec{p}, \vec{q}$ and $\vec{r}$ are three non-coplanar vectors in $ℝ^{3} .$ Let the components of a vector along $\vec{p}, \vec{q}$ and $\vec{r}$ be 4, 3 and 5, respectively. If the components of this vector $\vec{s}$ along $(- \vec{p} + \vec{q} + \vec{r}), (\vec{p} - \vec{q} + r), (- \vec{p} - \vec{q} + \vec{r})$ are $x, y$ and $z$ respectively, then the value of $2 x + y + z$ is

Difficult

Solution

Given $\bar{s} = 4 \bar{p} + 3 \bar{q} + 5 \bar{r}$ .
Let $\bar{s} = x (- \bar{p} + \bar{q} + \bar{r}) + y (\bar{p} - \bar{q} + \bar{r}) + z (- \bar{p} - \bar{q} + \bar{r})$
$\begin{array}{rcl} \Rightarrow & \bar{s} = (- x + y - z) \bar{p} + (x - y - z) \bar{q} + (x + y + z) \bar{r} \end{array}$
$\begin{array}{rcl} \Rightarrow & - x + y - z = 4, x - y - z = 3, x + y + z = 5 \end{array}$
$\begin{array}{rcl} S o l v i n g t h e a b o v e e q u a t i o n s, w e g e t \end{array}$
$\begin{array}{rcl} x = 4, y = \frac{9}{2}, z = - \frac{7}{2} ∴ 2 x + y + z = 9 \end{array}$

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