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  • Using vectors, find the value of k such that the points (k, – 10, 3), (1, –1, 3) and (3, 5, 3) are collinear.

Using vectors, find the value of k such that the points (k, – 10, 3), (1, –1, 3) and (3, 5, 3) are collinear.

Answers (1)

Let the points be A (k, -10, 3), B (1, -1, 3) and C (3, 5, 3).

Let us find the position vectors of these points.

Assume that O is the origin.

Position vector of A is given by,

\\ \overrightarrow{\mathrm{OA}}=\mathrm{ki}-10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ \text{Position vector of B is given by,} \\ \overrightarrow{\mathrm{OB}}=\hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}}\\ \text{Position vector of } \mathrm{C} \text{ is given by,}

\vec { OC } = 3 \hat { l } + 5 \hat { j } + 3 \hat { k }

Know that, two vectors are said to be collinear, if they lie on the same line or parallel lines.

Since, A (k, -10, 3), B (1, -1, 3) and C (3, 5, 3) are collinear, we can say that:

Sum of modulus of any two vectors will be equal to the modulus of third vector.

This means, we need to find | \vec { AB } |

To find : | \vec { AB } |

Position vector of B-Position vector of A

\Rightarrow \vec { AB } = \vec { OB } - \vec { OA }

\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{AB}}=(\hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}})-(\mathrm{k} \hat{\imath}-10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\\ &\Rightarrow \overrightarrow{\mathrm{AB}}=\hat{\imath}-k \hat{\imath}-\hat{\jmath}+10 \hat{\jmath}+3 \hat{\mathrm{k}}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{AB}}=(1-\mathrm{k}) \hat{\imath}+9 \hat{\mathrm{j}}\\ &\text { Now, }\\ &|\overrightarrow{\mathrm{AB}}|=\sqrt{(1-\mathrm{k})^{2}+9^{2}} \end{aligned}

\\ \begin{aligned} &\Rightarrow|\overrightarrow{\mathrm{AB}}|=\sqrt{(1-\mathrm{k})^{2}+81}_{\ldots(\mathrm{i})}\\ &\text { To find }|\overrightarrow{\mathrm{BC}}|_{:}\\ &\overrightarrow{\mathrm{BC}}=\text { position vector of C-Position vector of } \mathrm{B}\\ &\Rightarrow \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}} \end{aligned}

\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{BC}}=(3 \hat{\imath}+5 \hat{\jmath}+3 \hat{\mathrm{k}})-(\hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}})\\ &\Rightarrow \overrightarrow{\mathrm{BC}}=3 \hat{\imath}-\hat{\imath}+5 \hat{\jmath}+\hat{\jmath}+3 \hat{\mathrm{k}}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{BC}}=2 \hat{\imath}+6 \hat{j}\\ &\text { Now, }\\ &|\overrightarrow{\mathrm{BC}}|=\sqrt{2^{2}+6^{2}} \end{aligned}

\\ \begin{aligned} &\Rightarrow|\overrightarrow{\mathrm{BC}}|=\sqrt{4+36}\\ &\Rightarrow|\overrightarrow{\mathrm{BC}}|=\sqrt{40}\\ &\text { To find }\\ &|\overrightarrow{\mathrm{AC}}| \end{aligned}

\overrightarrow{\mathrm{AC}}  =  Position vector of C-Position vector of A

\\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}} \\ \Rightarrow \overrightarrow{\mathrm{AC}}=(3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})-(\mathrm{ki}-10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{AC}}=3 \hat{\mathrm{i}}-\mathrm{k} \hat{\mathrm{I}}+5 \hat{\mathrm{j}}+10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}-3 \hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AC}}=(3-\mathrm{k}) \hat{\mathrm{i}}+15 \hat{\mathrm{j}}

\\ \begin{aligned} &\text { Now, }\\ &|\overrightarrow{\mathrm{AC}}|=\sqrt{(3-\mathrm{k})^{2}+15^{2}}\\ &\Rightarrow|\overrightarrow{\mathrm{AC}}|=\sqrt{(3-\mathrm{k})^{2}+225}...(iii) \end{aligned}

\\ Take,\\ |\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{BC}}|=|\overrightarrow{\mathrm{AC}}|\\

Substitute values of  |\overrightarrow{\mathrm{AB}}|,|\overrightarrow{\mathrm{BC}}| \text { and }|\overrightarrow{\mathrm{AC}}|  from (i), (ii) and (iii) respectively. We get,

$$ \sqrt{(1-\mathrm{k})^{2}+81}+\sqrt{40}=\sqrt{(3-\mathrm{k})^{2}+225} $$\\

\\ Or\\ \Rightarrow \sqrt{(3-\mathrm{k})^{2}+225}-\sqrt{40}=\sqrt{(1-\mathrm{k})^{2}+81}\\ \Rightarrow \sqrt{\left(9+\mathrm{k}^{2}-6 \mathrm{k}\right)+225}-\sqrt{40}=\sqrt{\left(1+\mathrm{k}^{2}-2 \mathrm{k}\right)+81}\\ \left[\because\right. \text{by algebraic identity,} \left.(a-b)^{2}=a^{2}+b^{2}-2 a b\right]\\

\\ \Rightarrow \sqrt{\mathrm{k}^{2}-6 \mathrm{k}+225+9}-\sqrt{40}=\sqrt{\mathrm{k}^{2}-2 \mathrm{k}+81+1}\\ \Rightarrow \sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}-\sqrt{40}=\sqrt{\mathrm{k}^{2}-2 \mathrm{k}+82} \\ \text{Squaring on both sides,} \\\Rightarrow\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}-\sqrt{40}\right]^{2}=\left[\sqrt{\mathrm{k}^{2}-2 \mathrm{k}+82}\right]^{2}

\\ \begin{aligned} &\Rightarrow\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right]^{2}+[\sqrt{40}]^{2}-2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}]=\mathrm{k}^{2}-2 \mathrm{k}+82\\ &\left[\because \text { by algebraic identity, }(a-b)^{2}=a^{2}+b^{2}-2 a b\right]\\ &\Rightarrow \mathrm{k}^{2}-6 \mathrm{k}+234+40-2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}]=\mathrm{k}^{2}-2 \mathrm{k}+82\\ &\Rightarrow \mathrm{k}^{2}-6 \mathrm{k}+234+40-\mathrm{k}^{2}+2 \mathrm{k}-82=2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}] \end{aligned}

\\ \Rightarrow \mathrm{k}^{2}-\mathrm{k}^{2}-6 \mathrm{k}+2 \mathrm{k}+234+40-82=2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}] \\ \Rightarrow-4 \mathrm{k}+192=2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}] \\ \Rightarrow-4 \mathrm{k}+192=2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right] .2[\sqrt{10}] \\ \Rightarrow 4(-\mathrm{k}+48)=4\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{10}]

\\ \begin{aligned} &\Rightarrow-\mathrm{k}+48=\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234} \cdot \sqrt{10}\\ &\text { Again, squaring on both sides, we get }\\ &[48-\mathrm{k}]^{2}=\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234} \cdot \sqrt{10}\right]^{2}\\ &\Rightarrow(48)^{2}+k^{2}-2(48)(k)=\left(k^{2}-6 k+234\right)(10)\left[\because \text { by algebraic identity },(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \end{aligned}

\\ \begin{aligned} &\Rightarrow 2304+\mathrm{k}^{2}-96 \mathrm{k}=10 \mathrm{k}^{2}-60 \mathrm{k}+2340\\ &\Rightarrow 10 k^{2}-k^{2}-60 k+96 k+2340-2304=0\\ &\Rightarrow 9 \mathrm{k}^{2}+36 \mathrm{k}+36=0\\ &\Rightarrow 9\left(\mathrm{k}^{2}+4 \mathrm{k}+4\right)=0\\ &\Rightarrow \mathrm{k}^{2}+4 \mathrm{k}+4=0 \end{aligned}

\\ \Rightarrow \mathrm{k}^{2}+2 \mathrm{k}+2 \mathrm{k}+4=0 \\ \Rightarrow \mathrm{k}(\mathrm{k}+2)+2(\mathrm{k}+2)=0 \\ \Rightarrow(\mathrm{k}+2)(\mathrm{k}+2)=0 \\ \Rightarrow \mathrm{k}=-2 \text { or } \mathrm{k}=-2

Thus, value of k is -2.

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