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. λ x $\stackrel{\to }{\text{r}}$   = λ$\stackrel{\to }{\text{r}}$ |λ$\stackrel{\to }{\text{r}}$| = |λ| x | $\stackrel{\to }{\text{r}}$ | $\stackrel{\to }{\text{r}}$ and λ$\stackrel{\to }{\text{r}}$ are collinear. If λ = -1 then λ$\stackrel{\to }{\text{r}}$ = - $\stackrel{\to }{\text{r}}$ $\stackrel{\to }{\text{r}}$ + (- $\stackrel{\to }{\text{r}}$) = $\stackrel{\to }{\text{0}}$ If λ = $\frac{\text{1}}{\left|\stackrel{\to }{\text{r}}\right|}$ , $\stackrel{_}{\text{r}}$ ≠ $\stackrel{_}{\text{0}}$ λ$\stackrel{\to }{\text{r}}$ = $\frac{\text{1}}{\left|\stackrel{\to }{\text{r}}\right|}$ $\stackrel{\to }{\text{r}}$ | λ$\stackrel{\to }{\text{r}}$ | = $\frac{\text{1}}{\left|\stackrel{\to }{\text{r}}\right|}$ $\stackrel{\to }{\text{r}}$  = 1 ⇒ λ$\stackrel{\to }{\text{r}}$ = $\stackrel{^}{\text{r}}$ = $\frac{\text{1}}{\left|\stackrel{\to }{\text{r}}\right|}$ $\stackrel{\to }{\text{r}}$ For any two scalars, l and m, and any two vectors, $\stackrel{\to }{\text{a}}$ and : • l $\stackrel{\to }{\text{a}}$ + m = (l + m) • l(m$\stackrel{\to }{\text{a}}$) = (lm)$\stackrel{\to }{\text{a}}$ • l($\stackrel{\to }{\text{a}}$ + $\stackrel{\to }{\text{b}}$) = l $\stackrel{\to }{\text{a}}$ + l $\stackrel{\to }{\text{b}}$ 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. $\stackrel{\to }{\text{OQ}}$ = x$\stackrel{^}{\text{i}}$ $\stackrel{\to }{\text{OS}}$ = y$\stackrel{^}{\text{j}}$ $\stackrel{\to }{\text{OR}}$ = z$\stackrel{^}{\text{k}}$ From the figure: $\stackrel{\to }{\text{QA}}$ = $\stackrel{\to }{\text{OS}}$ = y$\stackrel{^}{\text{j}}$ In ∆ OQA: $\stackrel{\to }{\text{OA}}$ = $\stackrel{\to }{\text{OQ}}$ + $\stackrel{\to }{\text{QA}}$ [Triangle law of vector addition] ⇒ $\stackrel{\to }{\text{OA}}$ = x$\stackrel{^}{\text{i}}$ + y$\stackrel{^}{\text{j}}$ From the figure: $\stackrel{\to }{\text{AP}}$ = $\stackrel{\to }{\text{OR}}$ = z$\stackrel{^}{\text{k}}$ In ∆ OAP: $\stackrel{\to }{\text{OP}}$ = $\stackrel{\to }{\text{OA}}$ + $\stackrel{\to }{\text{AP}}$ [Triangle law of vector addition] ⇒ $\stackrel{\to }{\text{OP}}$ = x$\stackrel{^}{\text{i}}$ + y$\stackrel{^}{\text{j}}$ + z$\stackrel{^}{\text{k}}$ Component form of vector $\stackrel{\to }{\text{OP}}$ : $\stackrel{\to }{\text{OP}}$ = x$\stackrel{^}{\text{i}}$ + y$\stackrel{^}{\text{j}}$ + z$\stackrel{^}{\text{k}}$ In right angled ∆ OQA: ${\left|\stackrel{\to }{\text{OA}}\right|}^{\text{2}}$ = ${\left|\stackrel{\to }{\text{OQ}}\right|}^{\text{2}}$ + ${\left|\stackrel{\to }{\text{QA}}\right|}^{\text{2}}$ ⇒ ⃓${\left|\stackrel{\to }{\text{OA}}\right|}^{\text{2}}$ = x2 + y2 In right angled ∆ OAP: ⃓${\left|\stackrel{\to }{\text{OP}}\right|}^{\text{2}}$ = ${\left|\stackrel{\to }{\text{OA}}\right|}^{\text{2}}$ + ${\left|\stackrel{\to }{\text{AP}}\right|}^{\text{2}}$ ⇒ ⃓${\left|\stackrel{\to }{\text{OP}}\right|}^{\text{2}}$ = x2 + y2 + z2 ⇒ $\left|\stackrel{\to }{\text{OP}}\right|$ = $\sqrt{{\text{x}}^{\text{2}}\text{+}{\text{y}}^{\text{2}}\text{+}{\text{z}}^{\text{2}}}$ Direction cosines of $\stackrel{\to }{\text{OP}}$: l = cos α m = cos β n = cos γ Unit vector  $\stackrel{^}{\text{a}}$ in the direction of $\stackrel{\to }{\text{OP}}$: $\stackrel{^}{\text{a}}$ = l$\stackrel{^}{\text{i}}$ + m$\stackrel{^}{\text{j}}$ + n$\stackrel{^}{\text{k}}$ = cos α $\stackrel{^}{\text{i}}$ + cos β $\stackrel{^}{\text{j}}$ + cos ɣ $\stackrel{^}{\text{k}}$ $\stackrel{\to }{\text{a}}$ = a1$\stackrel{^}{\text{i}}$ + a2$\stackrel{^}{\text{j}}$ + a3$\stackrel{^}{\text{k}}$ λ $\stackrel{\to }{\text{a}}$ = λ a1$\stackrel{^}{\text{i}}$ + λ a2$\stackrel{^}{\text{j}}$ + λ a3$\stackrel{^}{\text{k}}$ $\stackrel{\to }{\text{b}}$ = b1$\stackrel{^}{\text{i}}$ + b2$\stackrel{^}{\text{j}}$ + b3$\stackrel{^}{\text{k}}$ $\stackrel{\to }{\text{a}}$ + $\stackrel{\to }{\text{b}}$ = (a1 + b1)$\stackrel{^}{\text{i}}$ + (a2 + b2)$\stackrel{^}{\text{j}}$ + (a3 + b3)$\stackrel{^}{\text{k}}$ $\stackrel{\to }{\text{a}}$ - $\stackrel{\to }{\text{b}}$ = (a1 – b1)$\stackrel{^}{\text{i}}$ + (a2 – b2)$\stackrel{^}{\text{j}}$ + (a3 – b3)$\stackrel{^}{\text{k}}$ $\stackrel{\to }{\text{a}}$ = $\stackrel{\to }{\text{b}}$ ⇔ a1 = b1, a2 = b2 and a3 = b3 $\stackrel{\to }{\text{a}}$ and $\stackrel{\to }{\text{b}}$ are collinear if only if there exists a scalar, λ, such that: a1$\stackrel{^}{\text{i}}$ + a2$\stackrel{^}{\text{j}}$ + a3$\stackrel{^}{\text{k}}$ = λ ( b1$\stackrel{^}{\text{i}}$ + b2$\stackrel{^}{\text{j}}$ + b3$\stackrel{^}{\text{k}}$ ) ⇒ a1$\stackrel{^}{\text{i}}$ + a2$\stackrel{^}{\text{j}}$ + a3$\stackrel{^}{\text{k}}$ =  λb1$\stackrel{^}{\text{i}}$ + λb2$\stackrel{^}{\text{j}}$ + λb3$\stackrel{^}{\text{k}}$ ) ⇒a1 = λb1, a2 = λb2 and a3 = λb3 ⇒ $\frac{{\text{a}}_{\text{1}}}{{\text{b}}_{\text{1}}}$ = $\frac{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}}$ = $\frac{{\text{a}}_{\text{3}}}{{\text{b}}_{\text{3}}}$ = λ Component form ofVector Joining Two Points Given A1 (x1, y1, z1) and A2 (x2, y2, z2). In ∆OP1P2: $\stackrel{\to }{{\text{OP}}_{\text{2}}}$ = $\stackrel{\to }{{\text{OP}}_{\text{1}}}$ + $\stackrel{\to }{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  [By triangle law of vector addition] ⇒ $\stackrel{\to }{{\text{OP}}_{\text{1}}}$ = $\stackrel{\to }{{\text{OP}}_{\text{2}}}$ – $\stackrel{\to }{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  ....(1) $\stackrel{\to }{{\text{OP}}_{\text{1}}}$ = x1$\stackrel{^}{\text{i}}$ + y1$\stackrel{^}{\text{j}}$ + z1$\stackrel{^}{\text{k}}$ $\stackrel{\to }{{\text{OP}}_{\text{2}}}$ = x2$\stackrel{^}{\text{i}}$ + y2$\stackrel{^}{\text{j}}$ + z2$\stackrel{^}{\text{k}}$ ⇒ $\stackrel{\to }{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$ = ( x2$\stackrel{^}{\text{i}}$ + y2$\stackrel{^}{\text{j}}$ + z2$\stackrel{^}{\text{k}}$ ) – (x1$\stackrel{^}{\text{i}}$ + y1$\stackrel{^}{\text{j}}$ + z1$\stackrel{^}{\text{k}}$ ) ⇒ $\stackrel{\to }{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  = (x2 – x1)$\stackrel{^}{\text{i}}$ + (y2 – y1)$\stackrel{^}{\text{j}}$ + (z2 – z1)$\stackrel{^}{\text{k}}$ |  $\stackrel{\to }{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  | = $\sqrt{{\left({\text{x}}_{2}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}{\text{x}}_{1}\right)}^{2}+{\left({\text{y}}_{2}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}{\text{y}}_{1}\right)}^{2}+{\left({\text{z}}_{2}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}{\text{z}}_{1}\right)}^{2}}$

#### Summary

If we multiply any vector by a scalar quantity we get another vector. λ x $\stackrel{\to }{\text{r}}$   = λ$\stackrel{\to }{\text{r}}$ |λ$\stackrel{\to }{\text{r}}$| = |λ| x | $\stackrel{\to }{\text{r}}$ | $\stackrel{\to }{\text{r}}$ and λ$\stackrel{\to }{\text{r}}$ are collinear. If λ = -1 then λ$\stackrel{\to }{\text{r}}$ = - $\stackrel{\to }{\text{r}}$ $\stackrel{\to }{\text{r}}$ + (- $\stackrel{\to }{\text{r}}$) = $\stackrel{\to }{\text{0}}$ If λ = $\frac{\text{1}}{\left|\stackrel{\to }{\text{r}}\right|}$ , $\stackrel{_}{\text{r}}$ ≠ $\stackrel{_}{\text{0}}$ λ$\stackrel{\to }{\text{r}}$ = $\frac{\text{1}}{\left|\stackrel{\to }{\text{r}}\right|}$ $\stackrel{\to }{\text{r}}$ | λ$\stackrel{\to }{\text{r}}$ | = $\frac{\text{1}}{\left|\stackrel{\to }{\text{r}}\right|}$ $\stackrel{\to }{\text{r}}$  = 1 ⇒ λ$\stackrel{\to }{\text{r}}$ = $\stackrel{^}{\text{r}}$ = $\frac{\text{1}}{\left|\stackrel{\to }{\text{r}}\right|}$ $\stackrel{\to }{\text{r}}$ For any two scalars, l and m, and any two vectors, $\stackrel{\to }{\text{a}}$ and : • l $\stackrel{\to }{\text{a}}$ + m = (l + m) • l(m$\stackrel{\to }{\text{a}}$) = (lm)$\stackrel{\to }{\text{a}}$ • l($\stackrel{\to }{\text{a}}$ + $\stackrel{\to }{\text{b}}$) = l $\stackrel{\to }{\text{a}}$ + l $\stackrel{\to }{\text{b}}$ 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. $\stackrel{\to }{\text{OQ}}$ = x$\stackrel{^}{\text{i}}$ $\stackrel{\to }{\text{OS}}$ = y$\stackrel{^}{\text{j}}$ $\stackrel{\to }{\text{OR}}$ = z$\stackrel{^}{\text{k}}$ From the figure: $\stackrel{\to }{\text{QA}}$ = $\stackrel{\to }{\text{OS}}$ = y$\stackrel{^}{\text{j}}$ In ∆ OQA: $\stackrel{\to }{\text{OA}}$ = $\stackrel{\to }{\text{OQ}}$ + $\stackrel{\to }{\text{QA}}$ [Triangle law of vector addition] ⇒ $\stackrel{\to }{\text{OA}}$ = x$\stackrel{^}{\text{i}}$ + y$\stackrel{^}{\text{j}}$ From the figure: $\stackrel{\to }{\text{AP}}$ = $\stackrel{\to }{\text{OR}}$ = z$\stackrel{^}{\text{k}}$ In ∆ OAP: $\stackrel{\to }{\text{OP}}$ = $\stackrel{\to }{\text{OA}}$ + $\stackrel{\to }{\text{AP}}$ [Triangle law of vector addition] ⇒ $\stackrel{\to }{\text{OP}}$ = x$\stackrel{^}{\text{i}}$ + y$\stackrel{^}{\text{j}}$ + z$\stackrel{^}{\text{k}}$ Component form of vector $\stackrel{\to }{\text{OP}}$ : $\stackrel{\to }{\text{OP}}$ = x$\stackrel{^}{\text{i}}$ + y$\stackrel{^}{\text{j}}$ + z$\stackrel{^}{\text{k}}$ In right angled ∆ OQA: ${\left|\stackrel{\to }{\text{OA}}\right|}^{\text{2}}$ = ${\left|\stackrel{\to }{\text{OQ}}\right|}^{\text{2}}$ + ${\left|\stackrel{\to }{\text{QA}}\right|}^{\text{2}}$ ⇒ ⃓${\left|\stackrel{\to }{\text{OA}}\right|}^{\text{2}}$ = x2 + y2 In right angled ∆ OAP: ⃓${\left|\stackrel{\to }{\text{OP}}\right|}^{\text{2}}$ = ${\left|\stackrel{\to }{\text{OA}}\right|}^{\text{2}}$ + ${\left|\stackrel{\to }{\text{AP}}\right|}^{\text{2}}$ ⇒ ⃓${\left|\stackrel{\to }{\text{OP}}\right|}^{\text{2}}$ = x2 + y2 + z2 ⇒ $\left|\stackrel{\to }{\text{OP}}\right|$ = $\sqrt{{\text{x}}^{\text{2}}\text{+}{\text{y}}^{\text{2}}\text{+}{\text{z}}^{\text{2}}}$ Direction cosines of $\stackrel{\to }{\text{OP}}$: l = cos α m = cos β n = cos γ Unit vector  $\stackrel{^}{\text{a}}$ in the direction of $\stackrel{\to }{\text{OP}}$: $\stackrel{^}{\text{a}}$ = l$\stackrel{^}{\text{i}}$ + m$\stackrel{^}{\text{j}}$ + n$\stackrel{^}{\text{k}}$ = cos α $\stackrel{^}{\text{i}}$ + cos β $\stackrel{^}{\text{j}}$ + cos ɣ $\stackrel{^}{\text{k}}$ $\stackrel{\to }{\text{a}}$ = a1$\stackrel{^}{\text{i}}$ + a2$\stackrel{^}{\text{j}}$ + a3$\stackrel{^}{\text{k}}$ λ $\stackrel{\to }{\text{a}}$ = λ a1$\stackrel{^}{\text{i}}$ + λ a2$\stackrel{^}{\text{j}}$ + λ a3$\stackrel{^}{\text{k}}$ $\stackrel{\to }{\text{b}}$ = b1$\stackrel{^}{\text{i}}$ + b2$\stackrel{^}{\text{j}}$ + b3$\stackrel{^}{\text{k}}$ $\stackrel{\to }{\text{a}}$ + $\stackrel{\to }{\text{b}}$ = (a1 + b1)$\stackrel{^}{\text{i}}$ + (a2 + b2)$\stackrel{^}{\text{j}}$ + (a3 + b3)$\stackrel{^}{\text{k}}$ $\stackrel{\to }{\text{a}}$ - $\stackrel{\to }{\text{b}}$ = (a1 – b1)$\stackrel{^}{\text{i}}$ + (a2 – b2)$\stackrel{^}{\text{j}}$ + (a3 – b3)$\stackrel{^}{\text{k}}$ $\stackrel{\to }{\text{a}}$ = $\stackrel{\to }{\text{b}}$ ⇔ a1 = b1, a2 = b2 and a3 = b3 $\stackrel{\to }{\text{a}}$ and $\stackrel{\to }{\text{b}}$ are collinear if only if there exists a scalar, λ, such that: a1$\stackrel{^}{\text{i}}$ + a2$\stackrel{^}{\text{j}}$ + a3$\stackrel{^}{\text{k}}$ = λ ( b1$\stackrel{^}{\text{i}}$ + b2$\stackrel{^}{\text{j}}$ + b3$\stackrel{^}{\text{k}}$ ) ⇒ a1$\stackrel{^}{\text{i}}$ + a2$\stackrel{^}{\text{j}}$ + a3$\stackrel{^}{\text{k}}$ =  λb1$\stackrel{^}{\text{i}}$ + λb2$\stackrel{^}{\text{j}}$ + λb3$\stackrel{^}{\text{k}}$ ) ⇒a1 = λb1, a2 = λb2 and a3 = λb3 ⇒ $\frac{{\text{a}}_{\text{1}}}{{\text{b}}_{\text{1}}}$ = $\frac{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}}$ = $\frac{{\text{a}}_{\text{3}}}{{\text{b}}_{\text{3}}}$ = λ Component form ofVector Joining Two Points Given A1 (x1, y1, z1) and A2 (x2, y2, z2). In ∆OP1P2: $\stackrel{\to }{{\text{OP}}_{\text{2}}}$ = $\stackrel{\to }{{\text{OP}}_{\text{1}}}$ + $\stackrel{\to }{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  [By triangle law of vector addition] ⇒ $\stackrel{\to }{{\text{OP}}_{\text{1}}}$ = $\stackrel{\to }{{\text{OP}}_{\text{2}}}$ – $\stackrel{\to }{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  ....(1) $\stackrel{\to }{{\text{OP}}_{\text{1}}}$ = x1$\stackrel{^}{\text{i}}$ + y1$\stackrel{^}{\text{j}}$ + z1$\stackrel{^}{\text{k}}$ $\stackrel{\to }{{\text{OP}}_{\text{2}}}$ = x2$\stackrel{^}{\text{i}}$ + y2$\stackrel{^}{\text{j}}$ + z2$\stackrel{^}{\text{k}}$ ⇒ $\stackrel{\to }{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$ = ( x2$\stackrel{^}{\text{i}}$ + y2$\stackrel{^}{\text{j}}$ + z2$\stackrel{^}{\text{k}}$ ) – (x1$\stackrel{^}{\text{i}}$ + y1$\stackrel{^}{\text{j}}$ + z1$\stackrel{^}{\text{k}}$ ) ⇒ $\stackrel{\to }{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  = (x2 – x1)$\stackrel{^}{\text{i}}$ + (y2 – y1)$\stackrel{^}{\text{j}}$ + (z2 – z1)$\stackrel{^}{\text{k}}$ |  $\stackrel{\to }{{{\text{P}}_{\text{1}}\text{P}}_{\text{2}}}$  | = $\sqrt{{\left({\text{x}}_{2}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}{\text{x}}_{1}\right)}^{2}+{\left({\text{y}}_{2}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}{\text{y}}_{1}\right)}^{2}+{\left({\text{z}}_{2}\mathrm{\text{}}\mathrm{\text{-}}\mathrm{\text{}}{\text{z}}_{1}\right)}^{2}}$

Next