1
TG EAPCET 2024 (Online) 10th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
If $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{c}=-\hat{\mathbf{k}}$ are position vectors of two points and $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}, \mathbf{d}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ are two vectors, then the lines $\mathbf{r}=\mathbf{a}+t \mathbf{b}, \mathbf{r}=\mathbf{c}+s \mathbf{d}$ are
A
skew lines, when $\lambda=\frac{19}{3}$
B
coplanar, $\forall \lambda \in R$
C
skew lines when $\lambda \neq \frac{19}{3}$
D
coplanar, when $\lambda \neq \frac{19}{3}$
2
TG EAPCET 2024 (Online) 10th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three vectors each having $\sqrt{2}$ magnitude such that $(\mathbf{a}, \mathbf{b})=(\mathbf{b}, \mathbf{c})=(\mathbf{c}, \mathbf{a})=\frac{\pi}{3}$. If $\mathbf{x}=\mathbf{a} \times(\mathbf{b} \times \mathbf{c})$ and $\mathbf{y}=\mathbf{b} \times(\mathbf{c} \times \mathbf{a})$, then
A
$|\mathbf{x}|=|y|$
B
$|x|=\sqrt{2}|y|$
C
$|x|=2|y|$
D
$|x|+|y|=2$
3
TG EAPCET 2024 (Online) 10th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

$\mathbf{a}$ 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

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

A
$|\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}|$
B
$|\mathbf{a}\|\mathbf{b}\| \mathbf{c}|$
C
$|a|^{2}+|b|^{2}+\mid d^{2}$
D
$|\mathbf{a}|^{2}|\mathbf{b}|^{2}|\mathbf{c}|^{2}$
4
TG EAPCET 2024 (Online) 10th May Evening Shift
MCQ (Single Correct Answer)
+1
-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 $\mathbf{a} \cdot(\mathbf{c} \times \mathbf{b}-\mathbf{b}-\mathbf{c})=$
A
32
B
24
C
20
D
36
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