Question

Let \vec{a}=\hat{i}+2 \hat{j}+4 \hat{k}, \vec{b}=\hat{i}+\lambda \hat{j}+4 \hat{k}
a
=
i
^
+2
j
^
​ +4
k
^
,
b
=
i
^
+λ
j
^
​ +4
k
^
and \vec{c}=2 \hat{i}+4 \hat{j}+\left(\lambda^{2}-1\right) \hat{k}
c
=2
i
^
+4
j
^
​ +(λ
2
−1)
k
^
be coplanar vectors \lambda \neq \pm 3 .λ≠±3. Then \vec{a} \cdot \vec{c}
a
⋅
c
is

Answer 1

$a\cdot c=22$

Step-by-step explanation:

$\vec{a}=\hat{i}+2\hat{j}+4\hat{k}$

$\vec{b}=\hat{i}+\lambda\hat{j}+4\hat{k}$

$\vec{c}=2\hat{i}+4\hat{j}+(\lambda^2-1)\hat{k}$

$\lambda\neq \pm 3$

We are given that vectors a,b and c are coplanar

Therefore,$a.(b\times c)=0$

$a\cdot(b\times c)=\begin{vmatrix}1&2&4\\1&\lambda&4\\2&4&(\lambda^2-1)\end{vmatrix}$

$a\cdot(b\times c)=\lambda(\lambda^2-1)-16-2(\lambda^2-1-8)+4(4-2\lambda)$

$a\cdot(b\times c)=\lambda^3-\lambda-16-2\lambda^2+18+16-8\lambda$

$a\cdot(b\times c)=\lambda^3-2\lambda^2-9\lambda+18=0$

$\lambda^2(\lambda-2)-9(\lambda-2)=0$

$(\lambda-2)(\lambda^2-9)=0$

$(\lambda-2)(\lambda+3)(\lambda-3)=0$

$\lambda-2=0\implies \lambda=2$

$\lambda+3=0\implies \lambda=-3$

$\lambda-3=0\implies \lambda=3$

But $\lambda\neq \pm 3$

Therefore, possible value of $\lambda=2$

Substitute the value then we get

$\vec{c}=2\hat{i}+4\hat{j}+3\hat{k}$

$a\cdot c=(\hat{i}+2\hat{j}+4\hat{k})\cdot (2\hat{i}+4\hat{j}+3\hat{k})$

$a\cdot c=2+8+12=22$

$i\cdot j=j\cdot k=k\cdot i=0,i\cdot i=j\cdot j=k\cdot k=1$

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