TS EAMCET 2020 (Online) 14th September Morning Shift | Vector Algebra Question 108 | Mathematics

If $\mathbf{a}=2 \mathbf{u}+3 \mathbf{v}+7 \mathbf{w}, b=\mathbf{u}+\mathbf{v}-2 \mathbf{w}$ and $\mathbf{c}=-\mathbf{u}-2 \mathbf{v}-3 \mathbf{w}$ then $\left|\frac{[\mathbf{u} \mathbf{v} \mathbf{w}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}\right|(\mathbf{a}+\mathbf{b}+\mathbf{c})=$

$12(\mathbf{u}+\mathbf{v}+\mathbf{w})$

$3(\mathbf{u}+\mathbf{v}+\mathbf{w})$

$\frac{2}{3}(\mathbf{u}+\mathbf{v}+\mathbf{w})$

$\frac{1}{3}(u+v+w)$

TS EAMCET 2020 (Online) 14th September Morning Shift

MCQ (Single Correct Answer)

-0

Let $\mathbf{V}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{W}=\hat{\mathbf{i}}+3 \hat{\mathbf{k}}$. If $\mathbf{U}$ is a unit vector, then the maximum value of $[\mathbf{U} \mathbf{V} \mathbf{W}]$ is

-1

$\sqrt{10}+\sqrt{16}$

$\sqrt{59}$

$\sqrt{60}$

TS EAMCET 2020 (Online) 11th September Evening Shift

MCQ (Single Correct Answer)

-0

The equation of the plane in normal form passing through the point $A(\bar{a})$, parallel to a vector $\bar{b}$ and containing a vector $\bar{c}$ is

$\mathbf{r} \cdot \frac{\mathbf{c} \times \mathbf{a}}{|\mathbf{c} \times \mathbf{a}|}=\left|\frac{\mathbf{a} \times \mathbf{b}}{\mathbf{a} \times \mathbf{c}}\right|$

$\mathbf{r} \cdot \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|}=\frac{[\mathbf{a} \mathbf{b c}]}{|\mathbf{b} \times \mathbf{c}|}$

$\mathbf{r} \cdot \frac{\mathbf{b} \times \mathbf{c}}{|\mathbf{b} \times \mathbf{c}|}=\frac{[\mathbf{a} \mathbf{b c}]}{|\mathbf{b} \times \mathbf{c}|}$

$\mathbf{r} \cdot[\mathbf{a} \mathbf{b c}] \mathbf{a}=\frac{|\mathbf{b} \times \mathbf{c}|}{|\mathbf{a} \times \mathbf{c}|}$

TS EAMCET 2020 (Online) 11th September Evening Shift

MCQ (Single Correct Answer)

-0

$\mathbf{x}, \mathbf{y}, \mathbf{z}$ are three vectors each of magnitude $\sqrt{2}$ and each making an angle $60^{\circ}$ with one another. If $\mathbf{a}=\mathbf{x} \times(\mathbf{y} \times \mathbf{z}), \mathbf{b}=\mathbf{y} \times(\mathbf{z} \times \mathbf{x}), \mathbf{c}=\mathbf{x} \times \mathbf{y}$, then $\mathbf{x}=$

$\frac{1}{2}[(\mathrm{a}+\mathrm{b}) \times \mathrm{c}-(\mathrm{a}+\mathrm{b})]$

$\frac{1}{2}[c+a-b]$

$\frac{1}{2}[(\mathbf{a}+\mathbf{b}) \times \mathbf{c}+(\mathbf{a}+\mathbf{b})]$

$\frac{1}{2}[(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}-\mathbf{a}+\mathbf{b}]$

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