Question

.5: If a = i - j , b = i+j+k and c be a vector such that a x c + b = o and a.c= 4, then Ic|^2
is equal to_

Answer 1

Products 1 - 8 · This brings us to the calculus of several variables. .... A+B= PR=C+D. Then V =+A+$Bequals W=+C++D. ... The combinations of i and j (or i, j, k) produce all vectors v.

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Answer 2

$\mid c\mid ^2=\frac{19}{2}$

Step-by-step explanation:

Let c=xi+yj+zk be the vector

$a=i-j$

$b=i+j+k$

$a\times c+b=0$

$a\cdot c=4$

$a\times c=\begin{vmatrix}i&j&k\\1&-1&0\\x&y&z\end{vmatrix}$

$a\times c=-zi-zj+(x+y)k$

$a\cdot c=(i-j)\cdot (xi+yj+zk)$

$x-y=4$..(1)

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

Substitute the values then we get

$a\times c+b=-zi-zj+(x+y)k+i+j+k=(1-z)i+(1-z)j+(x+y+1)k$

$a\times c+b=0$

$(1-z)i+(1-z)j+(x+y++1)=0i+oj+ok$

$1-z=0$

$z=1$

$x+y+1=0$

$x+y=-1$..(2)

Adding equation(1) and equation(2)

$2x=3$

$x=\frac{3}{2}$

Substitute the value of x in equation (1)

$\frac{3}{2}-y=4$

$y=\frac{3}{2}-4=\frac{3-8}{2}=-\frac{5}{2}$

Therefore, vector c=$\frac{3}{2}i-\frac{5}{2}j+k$

$\mid c\mid^2=x^2+y^2+z^2$

Where x=Coefficient of x

y=Coefficient of y

z= Coefficient of z

$\mid c\mid ^2=((\frac{3}{2})^2+(-\frac{5}{2})^2+1^2)=\frac{19}{2}$

$\mid c\mid ^2=\frac{19}{2}$

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https://brainly.in/question/9085777:answered by Spra