Q.Write a short note on the scalar product between two vectors.

Perpendicular vectors have dot product zero: $\bar a\cdot\bar b=0$.

$\bar a\cdot\bar b=(3)(1)+(2)(p)+(9)(3)=3+2p+27=2p+30$

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Use $|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta$ to find $\theta$, then $|\bar a\cdot\bar b|=|\bar a||\bar b|\cos\theta$.

Given $|\bar a|=5$, $|\bar b|=13$, $|\bar a\times\bar b|=25$.

$|\bar a\times\bar b|=|\bar a||\bar b|\sin\theta \implies 25 = 5\cdot13\cdot\sin\theta = 65\sin\theta$

$\sin\theta=\frac{25}{65}=\frac{5}{13}$

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$|\bar a+\bar b|=|\bar a-\bar b|$ forces $\bar a\cdot\bar b=0$, so the vectors are perpendicular.

Step 1 — Square both sides.

$|\bar a+\bar b|^2 = |\bar a-\bar b|^2$

$(\bar a+\bar b)\cdot(\bar a+\bar b) = (\bar a-\bar b)\cdot(\bar a-\bar b)$

$|\bar a|^2+2\bar a\cdot\bar b+|\bar b|^2 = |\bar a|^2-2\bar a\cdot\bar b+|\bar b|^2$

Step 2 — Simplify.

$4\,\bar a\cdot\bar b = 0 \implies \bar a\cdot\bar b = 0$

Step 3 — Interpret. …

Two vectors are perpendicular exactly when their dot product is zero; solving that equation gives $\lambda$.

Given $\bar{u} = \bar{i} - \lambda\bar{j} + 2\bar{k}$ and $\bar{v} = 8\bar{i} + 6\bar{j} - \bar{k}$.

Step 1. Vectors are at right angles iff $\bar{u}\cdot\bar{v} = 0$.

Step 2. Compute the dot product: …

Two vectors are perpendicular exactly when their dot product is zero — set the dot product to zero and solve the resulting quadratic in $\lambda$.

Given $\bar{u}=\lambda\bar{i}-3\bar{j}+5\bar{k}$ and $\bar{v}=2\lambda\bar{i}-\lambda\bar{j}-\bar{k}$, perpendicular means $\bar{u}\cdot\bar{v}=0$.

$\bar{u}\cdot\bar{v} = \lambda(2\lambda) + (-3)(-\lambda) + 5(-1) = 2\lambda^2 + 3\lambda - 5 = 0$

Solve using the quadratic formula: …