Let a=6 i+9 j+12 k, b= i+11 j-2 k and c be vectors such that a c=… - Vidyadip

MCQ

Let $\vec{a}=6 \hat{i}+9 \hat{j}+12 \hat{k}, \vec{b}=\alpha \hat{i}+11 \hat{j}-2 \hat{k}$ and $\vec{c}$ be vectors such that $\vec{a} \times \vec{c}=\vec{a} \times \vec{b}$. If $\vec{a} \cdot \vec{c}=-12$, $\vec{c} .(\hat{i}-2 \hat{j}+\hat{k})=5$, then $\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})$ is equal to $.............$.

A
$10$
✓
$11$
C
$12$
D
$13$

✓

Answer

Correct option: B.

$11$

b
$\overrightarrow{ a } \times \overrightarrow{ c }=\overrightarrow{ a } \times 5$

$\Rightarrow \overrightarrow{ a } \times(\overrightarrow{ c }-\overrightarrow{ b })=0$

$\vec{a} \|^{ I }(\overrightarrow{ c }-\overrightarrow{ b })$

$\therefore \overrightarrow{ a }=\lambda(\overrightarrow{ c }-\overrightarrow{ b })$

$(6,9,12)=\lambda[x-\alpha, y-11, z+2]$

$\frac{x-\alpha}{2}=\frac{y-11}{3}=\frac{z+2}{4}$

$4 y-44=3 z+6$

$4 y-3 z=50$

$6 x+9 y+12 z=-12$

$2 x+3 y+4 z=-4$

$(\because x-2 y+z=5)$

$2 x-4 y+2 z=10$

$+\quad-$

$7 y+2 z=-14 \ldots(2)$

$8 y-6 z=100$

$21 y+6 z=-42$

$29 y =58$

$y =2, z =-14$

$\therefore x -4-14=5$

$x =23$

$\overline{ c }=(23,2,-14)$

$\overline{ c } \cdot(1,1,1)=23+2-14=11$

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