TG EAPCET 2024 (Online) 10th May Evening Shift | Vector Algebra Question 6 | Mathematics

$\mathbf{a}$ is a vector perpendicular to the plane containing non zero vectors $\mathbf{b}$ and $\mathbf{c}$. If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are such that

$|\mathbf{a}+\mathbf{b}+\mathbf{c}|=\sqrt{|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+|\mathbf{c}|^{2}}$, then

$|(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}|+|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=$

$|\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}|$

$|\mathbf{a}\|\mathbf{b}\| \mathbf{c}|$

$|a|^{2}+|b|^{2}+\mid d^{2}$

$|\mathbf{a}|^{2}|\mathbf{b}|^{2}|\mathbf{c}|^{2}$

TG EAPCET 2024 (Online) 10th May Evening Shift

MCQ (Single Correct Answer)

-0

If $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=3(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})$ and $\mathbf{c}$ is a vector such that $\mathbf{a} \times \mathbf{c}=\mathbf{b}$ and $\mathbf{a} . \mathbf{c}=3$, then $\mathbf{a} \cdot(\mathbf{c} \times \mathbf{b}-\mathbf{b}-\mathbf{c})=$

TG EAPCET 2024 (Online) 10th May Morning Shift

MCQ (Single Correct Answer)

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$P$ and $Q$ are the points of trisection of the segment $A B$. If $2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-6 \hat{\mathbf{k}}$ are the position vectors of $A$ and $B$ respectively, then the position vector of the point which divides $P Q$ in the ratio $2: 3$ is

$\frac{1}{15}(44 \hat{\mathbf{i}}-33 \hat{\mathbf{j}}-18 \hat{\mathbf{k}})$

$\frac{1}{5}(36 \hat{\mathbf{i}}-26 \hat{\mathbf{j}}-18 \hat{\mathbf{k}})$

$\frac{1}{5}(3 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-9 \hat{\mathbf{k}})$

$\frac{1}{15}(-3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+9 \hat{\mathbf{k}})$

TG EAPCET 2024 (Online) 10th May Morning Shift

MCQ (Single Correct Answer)

-0

The position vector of the point of intersection of the line joining the points $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and the line joining the points $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-6 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-7 \hat{\mathbf{k}}$ is

$\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$

$4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}$

$\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$

$\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$

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