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

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*Q is to solve this question without using the property of dot product.

From dot product ( How I solved it )
$2(4)+3(-4)+8(α) = 0 $. From here , we get α = +$\frac{1}{2}$
I want to know more different ways of how can I solve this Q.
 A: You can use the Pythagorean theorem. If you add two vectors, you get $6i-j+(8+\alpha)k$. Then we have $36+1+(8+\alpha)^2=4+9+64+16+16+\alpha^2 \implies 16\alpha=8$
A: Since the two vectors are perpendicular, a less natural way would be think of Pythagoras theorem
Let $\vec {AB}=2i+3j+8k$ and  $\vec {AC}=-4i+4j+\alpha k$
Since $\vec {AB} \perp\vec{AC} $, Therefore we have $\vec{BC}=\vec{AC}-\vec{AB}=-6i+j+(\alpha-8)k$
Also $BC^2=AC^2+AB^2$
$(-6)^2+(1)^2+(\alpha-8)^2=(16+16+\alpha^2)+(4+9+64)$
$\alpha=-0.5$
A: $\newcommand{\uvec}[1]{\boldsymbol{\hat{\textbf{#1}}}}$
You can use cross product.
As we know $|\vec p × \vec q|=|\vec p| |\vec q| \sin{\theta} \tag{i}$ , where $\theta$ is the angle between vectors $\vec p$ and $\vec q$.
Let $\vec p= 2 \uvec i + 3\uvec j +8 \uvec k$ and $\vec q= 4 \uvec i - 4 \uvec j +\alpha \uvec k$
Given that $\vec p$ and $\vec q$ are perpendicular, $\theta= 90° \implies \sin{\theta}=1$
Now, $\vec p × \vec q=\begin{vmatrix}
\uvec i & \uvec j& \uvec k\\
2 & 3 & 8\\
4 & -4 & \alpha
\end{vmatrix}$
$=(3 \alpha +32) \uvec i - (2 \alpha -32) \uvec j +(-8-12)\uvec k$
$\therefore |\vec p ×\vec q|=\sqrt{(3 \alpha +32)^2  + (2 \alpha -32)^2+20^2}$
$$|\vec p|= \sqrt{2^2+3^2+8^2}=\sqrt{77}, |\vec q|= \sqrt{4^2+(-4)^2+{\alpha}^2}$$
Using (i) we get,
$\sqrt{(3 \alpha +32)^2  + (2 \alpha -32)^2+400}=\sqrt{77(32+{\alpha}^2)}$
Solve for $\alpha$
A: The concept of orthogonality(perpendicularity in a broader sense), is built upon inner products. More specifically, if $<a,b>$ is the inner product of 2 vectors $a$ and $b$, the two vectors are said to be orthogonal if and only if $<a,b> = 0$.
The dot product is just one example for an inner product. Vectors that are orthogonal under one inner product aren't necessarily orthogonal under another inner product. Therefore, I don't think orthogonality can really be thought of as independent of the choice of inner product. Since the dot product is the most commonly used inner product, I guess your answer is as good as it gets.
Of course, you could try showing that the angle between the 2 vectors is $90^{0}$, but then again, angles also depend on the choice of inner product.
Hope it helps!
