If a vector $2i+3j+8k$ is perpendicular to the vector $4j-4i+\alpha k$ then the value of α is? solve without using the property of dot product.

Bodonimhk
2022-10-13
Answered

If a vector $2i+3j+8k$ is perpendicular to the vector $4j-4i+\alpha k$ then the value of α is? solve without using the property of dot product.

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periasemdy

Answered 2022-10-14
Author has **15** answers

Since the two vectors are perpendicular, a less natural way would be think of Pythagoras theorem

Let $\overrightarrow{AB}=2i+3j+8k$ and $\overrightarrow{AC}=-4i+4j+\alpha k$

Since $\overrightarrow{AB}\perp \overrightarrow{AC}$, Therefore we have $\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=-6i+j+(\alpha -8)k$

Also $B{C}^{2}=A{C}^{2}+A{B}^{2}$

$(-6{)}^{2}+(1{)}^{2}+(\alpha -8{)}^{2}=(16+16+{\alpha}^{2})+(4+9+64)$

$\alpha =-0.5$

Let $\overrightarrow{AB}=2i+3j+8k$ and $\overrightarrow{AC}=-4i+4j+\alpha k$

Since $\overrightarrow{AB}\perp \overrightarrow{AC}$, Therefore we have $\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=-6i+j+(\alpha -8)k$

Also $B{C}^{2}=A{C}^{2}+A{B}^{2}$

$(-6{)}^{2}+(1{)}^{2}+(\alpha -8{)}^{2}=(16+16+{\alpha}^{2})+(4+9+64)$

$\alpha =-0.5$

4enevi

Answered 2022-10-15
Author has **5** answers

You can use the Pythagorean theorem. If you add two vectors, you get$6i-j+(8+\alpha )k$. Then we have

$36+1+(8+\alpha {)}^{2}=4+9+64+16+16+{\alpha}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}16\alpha =8$

$36+1+(8+\alpha {)}^{2}=4+9+64+16+16+{\alpha}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}16\alpha =8$

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