Question

show that the points A(1, 2,7), B(2, 6,3) and C(3, 10,-1) are collinear using vectors​

Answer 1

$\large\underline{\sf{Solution-}}$

Given coordinates are

➢ Coordinates of A is (1, 2, 7)

➢ Coordinates of B are (2, 6, 3)

➢ Coordinates of C are (3, 10, - 1)

So, it means

$\rm :\longmapsto\:\vec{OA} = \hat{i} + 2\hat{j} +7 \hat{k}$

$\rm :\longmapsto\:\vec{OB} = 2\hat{i} + 6\hat{j} +3 \hat{k}$

$\rm :\longmapsto\:\vec{OC} = 3\hat{i} + 10\hat{j} - \hat{k}$

Now, Consider

$\rm :\longmapsto\:\vec{AB}$

$\rm \:  =  \: \vec{OB} - \vec{OA}$

$\rm \:  =  \: 2\hat{i} + 6\hat{j} + 3\hat{k} - (\hat{i} + 2\hat{j} + 7\hat{k})$

$\rm \:  =  \: 2\hat{i} + 6\hat{j} + 3\hat{k} - \hat{i} - 2\hat{j} - 7\hat{k}$

$\rm \:  =  \: \hat{i} + 4\hat{j} - 4\hat{k}$

$\red{\rm\implies \:\vec{AB} =  \: \hat{i} + 4\hat{j} - 4\hat{k} - - - (1)}$

Now, Consider

$\rm :\longmapsto\:\vec{BC}$

$\rm \:  =  \:\vec{OC} - \vec{OB}$

$\rm \:  =  \: 3\hat{i} + 10\hat{j} - \hat{k} - (2\hat{i} + 6\hat{j} + 3\hat{k})$

$\rm \:  =  \: 3\hat{i} + 10\hat{j} - \hat{k} - 2\hat{i} - 6\hat{j} - 3\hat{k}$

$\rm \:  =  \: \hat{i} + 4\hat{j} - 4\hat{k}$

$\rm \:  =  \: \vec{AB}$

$\bf\implies \:\vec{AB} = \vec{BC}$

$\bf\implies \:\vec{AB} \: \parallel \: \vec{BC}$

$\bf\implies \:A, \: B, \: C \: are \: collinear$

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More to Know

$\rm :\longmapsto\:\boxed{\tt{ \vec{A}.\vec{B} = \vec{B}.\vec{A} \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \vec{A} \times \vec{B} = - \vec{B} \times \vec{A} \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \vec{A} \times \vec{B} = 0\rm\implies \:\vec{A} \: \parallel \: \vec{B} \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \vec{A} .\vec{B} = 0\rm\implies \:\vec{A} \: \perp \: \vec{B} \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \vec{A} \times \vec{A}= 0 \: }}$

Answer 2

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given coordinates are

➢ Coordinates of A is (1, 2, 7)

➢ Coordinates of B are (2, 6, 3)

➢ Coordinates of C are (3, 10, - 1)

So, it means

$\rm :\longmapsto\:\vec{OA} = \hat{i} + 2\hat{j} +7 \hat{k}$

$\rm :\longmapsto\:\vec{OB} = 2\hat{i} + 6\hat{j} +3 \hat{k}$

$\rm :\longmapsto\:\vec{OC} = 3\hat{i} + 10\hat{j} - \hat{k}$

Now, Consider

$\rm :\longmapsto\:\vec{AB}$

$\rm \:  =  \: \vec{OB} - \vec{OA}$

$\rm \:  =  \: 2\hat{i} + 6\hat{j} + 3\hat{k} - (\hat{i} + 2\hat{j} + 7\hat{k})$

$\rm \:  =  \: 2\hat{i} + 6\hat{j} + 3\hat{k} - \hat{i} - 2\hat{j} - 7\hat{k}$

$\rm \:  =  \: \hat{i} + 4\hat{j} - 4\hat{k}$

$\red{\rm\implies \:\vec{AB} =  \: \hat{i} + 4\hat{j} - 4\hat{k} - - - (1)}$

Now, Consider

$\rm :\longmapsto\:\vec{BC}$

$\rm \:  =  \:\vec{OC} - \vec{OB}$

$\rm \:  =  \: 3\hat{i} + 10\hat{j} - \hat{k} - (2\hat{i} + 6\hat{j} + 3\hat{k})$

$\rm \:  =  \: 3\hat{i} + 10\hat{j} - \hat{k} - 2\hat{i} - 6\hat{j} - 3\hat{k}$

$\rm \:  =  \: \hat{i} + 4\hat{j} - 4\hat{k}$

$\rm \:  =  \: \vec{AB}$

$\bf\implies \:\vec{AB} = \vec{BC}$

$\bf\implies \:\vec{AB} \: \parallel \: \vec{BC}$

$\bf\implies \:A, \: B, \: C \: are \: collinear$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to Know

$\rm :\longmapsto\:\boxed{\tt{ \vec{A}.\vec{B} = \vec{B}.\vec{A} \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \vec{A} \times \vec{B} = - \vec{B} \times \vec{A} \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \vec{A} \times \vec{B} = 0\rm\implies \:\vec{A} \: \parallel \: \vec{B} \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \vec{A} .\vec{B} = 0\rm\implies \:\vec{A} \: \perp \: \vec{B} \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \vec{A} \times \vec{A}= 0 \: }}$