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Find the scalar components and magnitude of the vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$

$\begin{array}{1 1} (A) \text{The scalar components of PQ are } (x_2-x_1),\:( y_2-y_1)\;and \;( z_2-z_1),\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} \\ \text{(B) The scalar components of PQ are} (x_1-x_2),\:( y_1-y_2) \;and \;( z_1-z_2),(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2 \\\text{(C) The scalar components of PQ are} x_1\: y_1\; and \;z_1,(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2 \\ \text{(D) The scalar components of PQ are} x_2\: y_2\; and \;z_2,(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2 \end{array} $

1 Answer

  • $\overrightarrow a=x_1\hat i+y_1\hat j+z_1\hat k$
  • The magnitude of a vector $|\overrightarrow a|=\sqrt{x_1^2+y_1^2+z_1^2}$
Step 1:
Let the position vectors of $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ be $\overrightarrow {OP}$ and $\overrightarrow {OQ}$
$\overrightarrow {OP}=x_1\hat i+y_1\hat j+z_1\hat k$
$\overrightarrow {OQ}=x_2\hat i+y_2\hat j+z_2\hat k$
Therefore $\overrightarrow {PQ}=\overrightarrow {OQ}-\overrightarrow {OP}$
$\quad\qquad\qquad=(x_2\hat i+y_2\hat j+z_2\hat k)-(x_1\hat i+y_1\hat j+z_1\hat k)$
$\quad\qquad\qquad=(x_2-x_1)\hat i+(y_2-y_1)\hat j+(z_2-z_1)\hat k$
Step 2:
The scalar components of vector $\overrightarrow {PQ}$ are $(x_2-x_1),(y_2-y_1)$ and $(z_2-z_1)$
answered May 23, 2013 by sreemathi.v

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