Notes On Multiplication of a Vector by a Scalar - CBSE Class 12 Maths
If we multiply any vector by a scalar quantity we get another vector. Î» x $\stackrel{â†’}{\text{r}}$   = Î»$\stackrel{â†’}{\text{r}}$ |Î»$\stackrel{â†’}{\text{r}}$| = |Î»| x | $\stackrel{â†’}{\text{r}}$ | $\stackrel{â†’}{\text{r}}$ and Î»$\stackrel{â†’}{\text{r}}$ are collinear. If Î» = -1 then Î»$\stackrel{â†’}{\text{r}}$ = - $\stackrel{â†’}{\text{r}}$ $\stackrel{â†’}{\text{r}}$ + (- $\stackrel{â†’}{\text{r}}$) = $\stackrel{â†’}{\text{0}}$ If Î» = $\frac{\text{1}}{\left|\stackrel{â†’}{\text{r}}\right|}$ , $\stackrel{_}{\text{r}}$ â‰  $\stackrel{_}{\text{0}}$ Î»$\stackrel{â†’}{\text{r}}$ = $\frac{\text{1}}{\left|\stackrel{â†’}{\text{r}}\right|}$ $\stackrel{â†’}{\text{r}}$ | Î»$\stackrel{â†’}{\text{r}}$ | = $\frac{\text{1}}{\left|\stackrel{â†’}{\text{r}}\right|}$ $\stackrel{â†’}{\text{r}}$  = 1 â‡’ Î»$\stackrel{â†’}{\text{r}}$ = $\stackrel{^}{\text{r}}$ = $\frac{\text{1}}{\left|\stackrel{â†’}{\text{r}}\right|}$ $\stackrel{â†’}{\text{r}}$ For any two scalars, l and m, and any two vectors, $\stackrel{â†’}{\text{a}}$ and : â€¢ l $\stackrel{â†’}{\text{a}}$ + m = (l + m) â€¢ l(m$\stackrel{â†’}{\text{a}}$) = (lm)$\stackrel{â†’}{\text{a}}$ â€¢ l($\stackrel{â†’}{\text{a}}$ + $\stackrel{â†’}{\text{b}}$) = l $\stackrel{â†’}{\text{a}}$ + l $\stackrel{â†’}{\text{b}}$ Let P be a point in space with coordinates x, y, z. Let us draw a perpendicular, PA, from point P on to the XOY plane. Let Q and S be the feet of the perpendiculars drawn from point A to the X and Y axes, respectively. Now, Q, S and R are the points on the X, Y and Z axes, respectively, such that OQ is equal to X, OS is equal to Y and OR is equal to Z. Now, vectors I, J and K are unit vectors along the X, Y and Z axes, respectively. $\stackrel{â†’}{\text{OQ}}$ = x$\stackrel{^}{\text{i}}$ $\stackrel{â†’}{\text{OS}}$ = y$\stackrel{^}{\text{j}}$ $\stackrel{â†’}{\text{OR}}$ = z$\stackrel{^}{\text{k}}$ From the figure: $\stackrel{â†’}{\text{QA}}$ = $\stackrel{â†’}{\text{OS}}$ = y$\stackrel{^}{\text{j}}$ In âˆ† OQA: $\stackrel{â†’}{\text{OA}}$ = $\stackrel{â†’}{\text{OQ}}$ + $\stackrel{â†’}{\text{QA}}$ [Triangle law of vector addition] â‡’ $\stackrel{â†’}{\text{OA}}$ = x$\stackrel{^}{\text{i}}$ + y$\stackrel{^}{\text{j}}$ From the figure: $\stackrel{â†’}{\text{AP}}$ = $\stackrel{â†’}{\text{OR}}$ = z$\stackrel{^}{\text{k}}$ In âˆ† OAP: $\stackrel{â†’}{\text{OP}}$ = $\stackrel{â†’}{\text{OA}}$ + $\stackrel{â†’}{\text{AP}}$ [Triangle law of vector addition] â‡’ $\stackrel{â†’}{\text{OP}}$ = x$\stackrel{^}{\text{i}}$ + y$\stackrel{^}{\text{j}}$ + z$\stackrel{^}{\text{k}}$ Component form of vector $\stackrel{â†’}{\text{OP}}$ : $\stackrel{â†’}{\text{OP}}$ = x$\stackrel{^}{\text{i}}$ + y$\stackrel{^}{\text{j}}$ + z$\stackrel{^}{\text{k}}$ In right angled âˆ† OQA: ${\left|\stackrel{â†’}{\text{OA}}\right|}^{\text{2}}$ = ${\left|\stackrel{â†’}{\text{OQ}}\right|}^{\text{2}}$ + ${\left|\stackrel{â†’}{\text{QA}}\right|}^{\text{2}}$ â‡’ âƒ“${\left|\stackrel{â†’}{\text{OA}}\right|}^{\text{2}}$ = x2 + y2 In right angled âˆ† OAP: âƒ“${\left|\stackrel{â†’}{\text{OP}}\right|}^{\text{2}}$ = ${\left|\stackrel{â†’}{\text{OA}}\right|}^{\text{2}}$ + ${\left|\stackrel{â†’}{\text{AP}}\right|}^{\text{2}}$ â‡’ âƒ“${\left|\stackrel{â†’}{\text{OP}}\right|}^{\text{2}}$ = x2 + y2 + z2 â‡’ $\left|\stackrel{â†’}{\text{OP}}\right|$ = $\sqrt{{\text{x}}^{\text{2}}\text{+}{\text{y}}^{\text{2}}\text{+}{\text{z}}^{\text{2}}}$ Direction cosines of $\stackrel{â†’}{\text{OP}}$: l = cos Î± m = cos Î² n = cos Î³ Unit vector  $\stackrel{^}{\text{a}}$ in the direction of $\stackrel{â†’}{\text{OP}}$: $\stackrel{^}{\text{a}}$ = l$\stackrel{^}{\text{i}}$ + m$\stackrel{^}{\text{j}}$ + n$\stackrel{^}{\text{k}}$ = cos Î± $\stackrel{^}{\text{i}}$ + cos Î² $\stackrel{^}{\text{j}}$ + cos É£ $\stackrel{^}{\text{k}}$ $\stackrel{â†’}{\text{a}}$ = a1$\stackrel{^}{\text{i}}$ + a2$\stackrel{^}{\text{j}}$ + a3$\stackrel{^}{\text{k}}$ Î» $\stackrel{â†’}{\text{a}}$ = Î» a1$\stackrel{^}{\text{i}}$ + Î» a2$\stackrel{^}{\text{j}}$ + Î» a3$\stackrel{^}{\text{k}}$ $\stackrel{â†’}{\text{b}}$ = b1$\stackrel{^}{\text{i}}$ + b2$\stackrel{^}{\text{j}}$ + b3$\stackrel{^}{\text{k}}$ $\stackrel{â†’}{\text{a}}$ + $\stackrel{â†’}{\text{b}}$ = (a1 + b1)$\stackrel{^}{\text{i}}$ + (a2 + b2)$\stackrel{^}{\text{j}}$ + (a3 + b3)$\stackrel{^}{\text{k}}$ $\stackrel{â†’}{\text{a}}$ - $\stackrel{â†’}{\text{b}}$ = (a1 â€“ b1)$\stackrel{^}{\text{i}}$ + (a2 â€“ b2)$\stackrel{^}{\text{j}}$ + (a3 â€“ b3)$\stackrel{^}{\text{k}}$ $\stackrel{â†’}{\text{a}}$ = $\stackrel{â†’}{\text{b}}$ â‡” a1 = b1, a2 = b2 and a3 = b3 $\stackrel{â†’}{\text{a}}$ and $\stackrel{â†’}{\text{b}}$ are collinear if only if there exists a scalar, Î», such that: a1$\stackrel{^}{\text{i}}$ + a2$\stackrel{^}{\text{j}}$ + a3$\stackrel{^}{\text{k}}$ = Î» ( b1$\stackrel{^}{\text{i}}$ + b2$\stackrel{^}{\text{j}}$ + b3$\stackrel{^}{\text{k}}$ ) â‡’ a1$\stackrel{^}{\text{i}}$ + a2$\stackrel{^}{\text{j}}$ + a3$\stackrel{^}{\text{k}}$ =  Î»b1$\stackrel{^}{\text{i}}$ + Î»b2$\stackrel{^}{\text{j}}$ + Î»b3$\stackrel{^}{\text{k}}$ ) â‡’a1 = Î»b1, a2 = Î»b2 and a3 = Î»b3 â‡’ $\frac{{\text{a}}_{\text{1}}}{{\text{b}}_{\text{1}}}$ = $\frac{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}}$ = $\frac{{\text{a}}_{\text{3}}}{{\text{b}}_{\text{3}}}$ = Î» Component form ofVector Joining Two Points Given A1 (x1, y1, z1) and A2 (x2, y2, z2). In âˆ†OP1P2: $\stackrel{â†’}{{\text{OP}}_{\text{2}}}$ = $\stackrel{â†’}{{\text{OP}}_{\text{1}}}$ + $\stackrel{â†’}{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  [By triangle law of vector addition] â‡’ $\stackrel{â†’}{{\text{OP}}_{\text{1}}}$ = $\stackrel{â†’}{{\text{OP}}_{\text{2}}}$ â€“ $\stackrel{â†’}{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  ....(1) $\stackrel{â†’}{{\text{OP}}_{\text{1}}}$ = x1$\stackrel{^}{\text{i}}$ + y1$\stackrel{^}{\text{j}}$ + z1$\stackrel{^}{\text{k}}$ $\stackrel{â†’}{{\text{OP}}_{\text{2}}}$ = x2$\stackrel{^}{\text{i}}$ + y2$\stackrel{^}{\text{j}}$ + z2$\stackrel{^}{\text{k}}$ â‡’ $\stackrel{â†’}{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$ = ( x2$\stackrel{^}{\text{i}}$ + y2$\stackrel{^}{\text{j}}$ + z2$\stackrel{^}{\text{k}}$ ) â€“ (x1$\stackrel{^}{\text{i}}$ + y1$\stackrel{^}{\text{j}}$ + z1$\stackrel{^}{\text{k}}$ ) â‡’ $\stackrel{â†’}{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  = (x2 â€“ x1)$\stackrel{^}{\text{i}}$ + (y2 â€“ y1)$\stackrel{^}{\text{j}}$ + (z2 â€“ z1)$\stackrel{^}{\text{k}}$ |  $\stackrel{â†’}{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  | = $\sqrt{{\left({\text{x}}_{2}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}{\text{x}}_{1}\right)}^{2}+{\left({\text{y}}_{2}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}{\text{y}}_{1}\right)}^{2}+{\left({\text{z}}_{2}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}{\text{z}}_{1}\right)}^{2}}$

#### Summary

If we multiply any vector by a scalar quantity we get another vector. Î» x $\stackrel{â†’}{\text{r}}$   = Î»$\stackrel{â†’}{\text{r}}$ |Î»$\stackrel{â†’}{\text{r}}$| = |Î»| x | $\stackrel{â†’}{\text{r}}$ | $\stackrel{â†’}{\text{r}}$ and Î»$\stackrel{â†’}{\text{r}}$ are collinear. If Î» = -1 then Î»$\stackrel{â†’}{\text{r}}$ = - $\stackrel{â†’}{\text{r}}$ $\stackrel{â†’}{\text{r}}$ + (- $\stackrel{â†’}{\text{r}}$) = $\stackrel{â†’}{\text{0}}$ If Î» = $\frac{\text{1}}{\left|\stackrel{â†’}{\text{r}}\right|}$ , $\stackrel{_}{\text{r}}$ â‰  $\stackrel{_}{\text{0}}$ Î»$\stackrel{â†’}{\text{r}}$ = $\frac{\text{1}}{\left|\stackrel{â†’}{\text{r}}\right|}$ $\stackrel{â†’}{\text{r}}$ | Î»$\stackrel{â†’}{\text{r}}$ | = $\frac{\text{1}}{\left|\stackrel{â†’}{\text{r}}\right|}$ $\stackrel{â†’}{\text{r}}$  = 1 â‡’ Î»$\stackrel{â†’}{\text{r}}$ = $\stackrel{^}{\text{r}}$ = $\frac{\text{1}}{\left|\stackrel{â†’}{\text{r}}\right|}$ $\stackrel{â†’}{\text{r}}$ For any two scalars, l and m, and any two vectors, $\stackrel{â†’}{\text{a}}$ and : â€¢ l $\stackrel{â†’}{\text{a}}$ + m = (l + m) â€¢ l(m$\stackrel{â†’}{\text{a}}$) = (lm)$\stackrel{â†’}{\text{a}}$ â€¢ l($\stackrel{â†’}{\text{a}}$ + $\stackrel{â†’}{\text{b}}$) = l $\stackrel{â†’}{\text{a}}$ + l $\stackrel{â†’}{\text{b}}$ Let P be a point in space with coordinates x, y, z. Let us draw a perpendicular, PA, from point P on to the XOY plane. Let Q and S be the feet of the perpendiculars drawn from point A to the X and Y axes, respectively. Now, Q, S and R are the points on the X, Y and Z axes, respectively, such that OQ is equal to X, OS is equal to Y and OR is equal to Z. Now, vectors I, J and K are unit vectors along the X, Y and Z axes, respectively. $\stackrel{â†’}{\text{OQ}}$ = x$\stackrel{^}{\text{i}}$ $\stackrel{â†’}{\text{OS}}$ = y$\stackrel{^}{\text{j}}$ $\stackrel{â†’}{\text{OR}}$ = z$\stackrel{^}{\text{k}}$ From the figure: $\stackrel{â†’}{\text{QA}}$ = $\stackrel{â†’}{\text{OS}}$ = y$\stackrel{^}{\text{j}}$ In âˆ† OQA: $\stackrel{â†’}{\text{OA}}$ = $\stackrel{â†’}{\text{OQ}}$ + $\stackrel{â†’}{\text{QA}}$ [Triangle law of vector addition] â‡’ $\stackrel{â†’}{\text{OA}}$ = x$\stackrel{^}{\text{i}}$ + y$\stackrel{^}{\text{j}}$ From the figure: $\stackrel{â†’}{\text{AP}}$ = $\stackrel{â†’}{\text{OR}}$ = z$\stackrel{^}{\text{k}}$ In âˆ† OAP: $\stackrel{â†’}{\text{OP}}$ = $\stackrel{â†’}{\text{OA}}$ + $\stackrel{â†’}{\text{AP}}$ [Triangle law of vector addition] â‡’ $\stackrel{â†’}{\text{OP}}$ = x$\stackrel{^}{\text{i}}$ + y$\stackrel{^}{\text{j}}$ + z$\stackrel{^}{\text{k}}$ Component form of vector $\stackrel{â†’}{\text{OP}}$ : $\stackrel{â†’}{\text{OP}}$ = x$\stackrel{^}{\text{i}}$ + y$\stackrel{^}{\text{j}}$ + z$\stackrel{^}{\text{k}}$ In right angled âˆ† OQA: ${\left|\stackrel{â†’}{\text{OA}}\right|}^{\text{2}}$ = ${\left|\stackrel{â†’}{\text{OQ}}\right|}^{\text{2}}$ + ${\left|\stackrel{â†’}{\text{QA}}\right|}^{\text{2}}$ â‡’ âƒ“${\left|\stackrel{â†’}{\text{OA}}\right|}^{\text{2}}$ = x2 + y2 In right angled âˆ† OAP: âƒ“${\left|\stackrel{â†’}{\text{OP}}\right|}^{\text{2}}$ = ${\left|\stackrel{â†’}{\text{OA}}\right|}^{\text{2}}$ + ${\left|\stackrel{â†’}{\text{AP}}\right|}^{\text{2}}$ â‡’ âƒ“${\left|\stackrel{â†’}{\text{OP}}\right|}^{\text{2}}$ = x2 + y2 + z2 â‡’ $\left|\stackrel{â†’}{\text{OP}}\right|$ = $\sqrt{{\text{x}}^{\text{2}}\text{+}{\text{y}}^{\text{2}}\text{+}{\text{z}}^{\text{2}}}$ Direction cosines of $\stackrel{â†’}{\text{OP}}$: l = cos Î± m = cos Î² n = cos Î³ Unit vector  $\stackrel{^}{\text{a}}$ in the direction of $\stackrel{â†’}{\text{OP}}$: $\stackrel{^}{\text{a}}$ = l$\stackrel{^}{\text{i}}$ + m$\stackrel{^}{\text{j}}$ + n$\stackrel{^}{\text{k}}$ = cos Î± $\stackrel{^}{\text{i}}$ + cos Î² $\stackrel{^}{\text{j}}$ + cos É£ $\stackrel{^}{\text{k}}$ $\stackrel{â†’}{\text{a}}$ = a1$\stackrel{^}{\text{i}}$ + a2$\stackrel{^}{\text{j}}$ + a3$\stackrel{^}{\text{k}}$ Î» $\stackrel{â†’}{\text{a}}$ = Î» a1$\stackrel{^}{\text{i}}$ + Î» a2$\stackrel{^}{\text{j}}$ + Î» a3$\stackrel{^}{\text{k}}$ $\stackrel{â†’}{\text{b}}$ = b1$\stackrel{^}{\text{i}}$ + b2$\stackrel{^}{\text{j}}$ + b3$\stackrel{^}{\text{k}}$ $\stackrel{â†’}{\text{a}}$ + $\stackrel{â†’}{\text{b}}$ = (a1 + b1)$\stackrel{^}{\text{i}}$ + (a2 + b2)$\stackrel{^}{\text{j}}$ + (a3 + b3)$\stackrel{^}{\text{k}}$ $\stackrel{â†’}{\text{a}}$ - $\stackrel{â†’}{\text{b}}$ = (a1 â€“ b1)$\stackrel{^}{\text{i}}$ + (a2 â€“ b2)$\stackrel{^}{\text{j}}$ + (a3 â€“ b3)$\stackrel{^}{\text{k}}$ $\stackrel{â†’}{\text{a}}$ = $\stackrel{â†’}{\text{b}}$ â‡” a1 = b1, a2 = b2 and a3 = b3 $\stackrel{â†’}{\text{a}}$ and $\stackrel{â†’}{\text{b}}$ are collinear if only if there exists a scalar, Î», such that: a1$\stackrel{^}{\text{i}}$ + a2$\stackrel{^}{\text{j}}$ + a3$\stackrel{^}{\text{k}}$ = Î» ( b1$\stackrel{^}{\text{i}}$ + b2$\stackrel{^}{\text{j}}$ + b3$\stackrel{^}{\text{k}}$ ) â‡’ a1$\stackrel{^}{\text{i}}$ + a2$\stackrel{^}{\text{j}}$ + a3$\stackrel{^}{\text{k}}$ =  Î»b1$\stackrel{^}{\text{i}}$ + Î»b2$\stackrel{^}{\text{j}}$ + Î»b3$\stackrel{^}{\text{k}}$ ) â‡’a1 = Î»b1, a2 = Î»b2 and a3 = Î»b3 â‡’ $\frac{{\text{a}}_{\text{1}}}{{\text{b}}_{\text{1}}}$ = $\frac{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}}$ = $\frac{{\text{a}}_{\text{3}}}{{\text{b}}_{\text{3}}}$ = Î» Component form ofVector Joining Two Points Given A1 (x1, y1, z1) and A2 (x2, y2, z2). In âˆ†OP1P2: $\stackrel{â†’}{{\text{OP}}_{\text{2}}}$ = $\stackrel{â†’}{{\text{OP}}_{\text{1}}}$ + $\stackrel{â†’}{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  [By triangle law of vector addition] â‡’ $\stackrel{â†’}{{\text{OP}}_{\text{1}}}$ = $\stackrel{â†’}{{\text{OP}}_{\text{2}}}$ â€“ $\stackrel{â†’}{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  ....(1) $\stackrel{â†’}{{\text{OP}}_{\text{1}}}$ = x1$\stackrel{^}{\text{i}}$ + y1$\stackrel{^}{\text{j}}$ + z1$\stackrel{^}{\text{k}}$ $\stackrel{â†’}{{\text{OP}}_{\text{2}}}$ = x2$\stackrel{^}{\text{i}}$ + y2$\stackrel{^}{\text{j}}$ + z2$\stackrel{^}{\text{k}}$ â‡’ $\stackrel{â†’}{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$ = ( x2$\stackrel{^}{\text{i}}$ + y2$\stackrel{^}{\text{j}}$ + z2$\stackrel{^}{\text{k}}$ ) â€“ (x1$\stackrel{^}{\text{i}}$ + y1$\stackrel{^}{\text{j}}$ + z1$\stackrel{^}{\text{k}}$ ) â‡’ $\stackrel{â†’}{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  = (x2 â€“ x1)$\stackrel{^}{\text{i}}$ + (y2 â€“ y1)$\stackrel{^}{\text{j}}$ + (z2 â€“ z1)$\stackrel{^}{\text{k}}$ |  $\stackrel{â†’}{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  | = $\sqrt{{\left({\text{x}}_{2}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}{\text{x}}_{1}\right)}^{2}+{\left({\text{y}}_{2}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}{\text{y}}_{1}\right)}^{2}+{\left({\text{z}}_{2}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}{\text{z}}_{1}\right)}^{2}}$

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