Question 104: Motion in 2D & Vectors

Question

The three vector $\vec{A}=3 \hat{i}-2 \hat{j}+\hat{k}, \vec{B}=\hat{i}-3 \hat{j}+5 k$ and $\vec{C}=2 \hat{i}-\hat{j}+4 \hat{k}$ will form

Accepted Answer

$$ \begin{aligned} & \overrightarrow{\mathrm{A}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \ & \overrightarrow{\mathrm{~B}}=\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \mathrm{k} \ & \overrightarrow{\mathrm{C}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \ & |\mathrm{~A}|=\sqrt{3^2+(-2)^2+1^2}=\sqrt{9+4+1}=\sqrt{14} \ & |\mathrm{~B}|=\sqrt{1^2+(-3)^2+5^2}=\sqrt{1+9+25}=\sqrt{35} \ & |\mathrm{C}|=\sqrt{2^2+(-1)^2+4^2}=\sqrt{4+1+16}=\sqrt{21} \end{aligned} $$
$	herefore \quad$ The three vectors do not form an equilateral or isosceles triangle because none of the sides have equal lengths.
$$ \begin{aligned} \vec{A} \cdot \vec{B} & =(3)(1)+(-2)(-3)+(1)(5) \ & =3+6+5=14 \end{aligned} $$
$$ \begin{aligned} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{C}} & =(1)(2)+(-3)(-1)+(5)(4) \ & =2+3+20=25 \ \overrightarrow{\mathrm{~A}} \cdot \overrightarrow{\mathrm{C}} & =(3)(2)+(-2)(-1)+(1)(4) \ & =6+2+4=12 \end{aligned} $$
None of the sides of the triangle are perpendicular to each other, as the dot product of any pair of vectors is not zero.
If the vectors are coplanar, then they will form a triangle (unless any one of them is parallel to each other)
For three vectors to be coplanar, scalar triple product should be zero.
$$ A \cdot(B 	imes C)=\left|\begin{array}{lll} A_x & A_y & A_z \ B_x & B_y & B_z \ C_x & C_y & C_z \end{array}ight|=\left|\begin{array}{ccc} 3 & -2 & 1 \ 1 & -3 & 5 \ 2 & -1 & 4 \end{array}ight|=-28 $$
Since the scalar triple product is non-zero ( -28 ), the vectors $\mathrm{A}, \mathrm{B}$, and C are not coplanar and hence do not form a triangle.

Answer

isosceles triangle.

Answer

equilateral triangle.

Answer

no triangle.

Answer

right angled triangle.

Motion in 2D & Vectors

The three vector \vec{A}=3 \hat{i}-2 \hat{j}+\hat{k}, \vec{B}=\hat{i}-3 \hat{j}+5 k and \vec{C}=2 \hat{i}-\hat{j}+4 \hat{k} will form

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